Populating A Drop Box With Results From Database
I have a MySQL database with a table (category) with two fields, catId (int) and category (char(50)). What I want to do is to get all category names in this database and place all of them into a dropdown box on a web page so that the user can choose from the list of available categories.
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Populating Form From Database, Then Passing Results To Next Page
I have a multiple select input in a form that's being populated by a row from my database as such: <input type="checkbox" name="subm[]" value="$row[ID]"> That part is working fine as I can check the displayed page using View Source and see that the value is the correct row number from the database. It is then being submitted on a form by $_POST method to another page where I want to evaluate the checkboxes and display the contents of the entire row that corresponds to each value="$row[ID]" that have been checked. But I can't seem to get it to work. I'm having a problem passing the selected value. Can someone point me in the right direction? $query = ("SELECT * FROM `table`"); $result = mysql_query($query); print "<p>Data for Selections:"; print "<table border=2><tr><th>You chose:"; foreach ($_POST['subm'] as $value) { print "<tr><td>"; print "$row[ID]; "; print mysql_field_name($result, 1) . ": " . $row[name]."<br>"; print mysql_field_name($result, 2) . ": " . $row[address]."<br>"; print mysql_field_name($result, 3) . ": " . $row[city]."<br>"; print "</td></tr>"; print "</table> "; } if (!isset($_POST['subm'])){ print "<p>No matching entry "; } mysql_close();
Populating A Drop Down Menu
How would I go about populating a Drop Down menu from data in MySQL? If you can show me wither a URL to learn this, or if you feel like telling me here, that would be great.
Populating Drop Down And Table
I have a database with the following fields. Name | Company | Date values in each column could be repeated, or not. as in there could be several same names with the same company with different dates, or different names with same company. How do I populate a table with this info and have drop down boxes, so that I can narrow down the search? For example: Name | Company | Date 12 | 1 | 1929 13 | 1 | 1929 14 | 1 | 1929 12 | 2 | 1929 12 | 4 | 1929 13 | 1 | 1941 12 | 6 | 1929 So if in the drop down I select '12' under name, only those entries with 12 are shown, and then I can further sort it by selecting only '1' under Company. Hope I'm clear, I manage to complicate things when I post them.
Auto Populating A Drop Down Box
Basically I'm setting up a website which needs an populated drop down box made up from all fields in a specific column of a table in a mysql db.... Here's the code I've made up using various tutorials.... <?php $user = ""; $host = "" $password = "" $dbName = "" /* make connection to database */ mysql_connect($host, $user, $password) OR DIE( "Unable to connect to database"); mysql_select_db($dbName); //did you forget this line? $sql = "SELECT model FROM usedVehicles"; $query = mysql_query($sql) or die(mysql_error() . "<br>$sql"); //use the or die(...) part ONLY IN DEVELOPMENT ?> <form action="action" method="post"> <select name="option"> <?php while ($row = mysql_fetch_array($result)) { echo "<option value="" . $row['model'] . "">" . $row['model'] . "</option> "; } ?> </select> <input type="submit"> </form> I've left out the connection details for obvious reasons... When I upload and try to test this, jus a blank drop down appears... there are definately fields in the column as I have tried the query on phpMyAdmin.
Auto Populating Multiple Drop Downs.
im trying to create two dropdowns, i need the first one to be the category and the second one to be the subcategory. The category drop down autopopulates with the correct info from the database. and uses the table "category", the value of each drop down is represented by the "cat" field in the table (cat is basically and integer id number) and "Category" is used as what the user actually sees in the drop down (category is the actual word of the category). Once the category is selected i would like to have the sub category auto populate with everything that has the same values as the selected category (cat) Here is a break down of how the tables work. Table 1 Name: "category" Fields for Table 1: "cat" (the id number), "category" ( the actual name of the category) Table 2 Name: "subcategory" Fields for Table 2: "cat" (corresponds with the cat id from table 1 to pull the correct data), "subc" (the basic id of the subcategory), "subcat" the actual name of the subcategory. so the way i see it, have a normal drop down populated by a php query. then on change, populate subcategory drop down where cat = cat and display sub category.
Populating The Drop Down And Pressing The Submit Button
1. The first is I have two drop down menus. The first is "year" and the second is "mfr". When a user selects a year from the drop down it then populates the second drop down, mfr, from the MySQL database. Theat is working fine. But the problem I am having is the "submit" button (which I have labeled as "browse"). When I click it. Nothing happens, no action tacks place. I have looked over the code and I can't figure it out. (See Code Box 1 Below). 2. Right now the "mfr" drop down is populated by the MySQL database and reads with a list like "Acr", "Alp", etc. These are abreviations. I need to set up an array to have them instead read the entire mfr name. Example: Instead of "Acr" it needs to be "Acura". Instead of "Alp" it needs to be "Alpine". I need these full names to appear in the drop down. I know I need to do something like this (See Code Box 2 Below) but I can't get my finger on it. Code:
Populating City,state,country Drop Down Menu
In the registration form I have city, state, country fields. I was wondering if there was a database available on the net which has the list of states in each of the countries. That way when a user selects a country I could automatically populate the state drop down menu ...
Populating Array With Mysql Results
I'm returning a result set of one field in a table, and want to populate an array with the results. I can't for the life of me figure out a simple way to do this without using mysql_fetch_array() to cycle through the results, append that to a var, then explode that into a var and pass that. PHP Code:
Populating A Select Box From A Database
I have a subscription database that allows you to edit a subscribers data. For example, you enter a search term like "Bob" and it returns a list of everyone named Bob. You click the one you want and it goes to a subscriber detail page, where all of his data is populated to a form. You can directly edit this form and click save to overwrite it. The only issue is the select boxes (drop downs) - say I have four options in the drop down box, red, blue, green and black. On my details page for bob I end up with five, I have those four PLUS whatever was saved for his color, so it shows up twice. If Bob is green, my select box looks like: Green Red Blue Green Black How can I make it so that Green only shows up once?
Populating A Form Wth Database Info
I've coded a piece of code which populates a form with data read from the database: $connection=mysql_connec ("localhost", "f2821842", "f2821842"); $result=mysql_select_db("QUERIES"); $query=mysql_query("Select * from Emp_Details where emp_num = '$employnum'"); while($row=mysql_fetch_array($query)) {$empname=$row['emp_name'];} <form name="webregform" action="webadmin2.php" method="post"> <input name="requiredname" type="text" size="30" value="<?php echo "$empname"; ?>"> </form> When I echo the $requiredname, I get spaces and no data, and I know that $empname is not a space-it does read a value in a database. 1. How can I get $requiredname to print a value?
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I'm trying to create a drop-down of the 'members' column of the 'roster' table. As you'll see it looks completly ****ed up and in no way works. PHP Code:
Drop Down Populated By Database
Hopefully someone can help me with this: I currently have a php file that pulls a query from the field "category" from my database. this file is called category.php. I am trying to populate a drop down box in another page (called add_subpage.php) from the results of the category page. the code below is on the add_subpage.php file: (It was my best guess and does not work) <select name="category"> <option value="CAT"<?php include 'category.php' ?>></option> </select> Can anyone tell me how to make this drop down populate from the category php file?
Drop Down Populated By Database
I currently have a php file that pulls a query from the field "category" from my database. this file is called category.php. I am trying to populate a drop down box in another page (called add_subpage.php) from the results of the category page. the code below is on the add_subpage.php file: (It was my best guess and does not work) <select name="category"> <option value="CAT"<?php include 'category.php' ?>></option</select> Can anyone tell me how to make this drop down populate from the category php file?
Drop Down With Assigned Value From Database
I have a database that has customers in it. When they purchase things from me, each purchase is either "online" or "onsite". So a selection must be made as either "online" or "onsite" when the item is entered into the database. Often, I have to update this value for an item later on from "online" to "onsite" and vice versa. Is there anyway to populate the "Type" field with whatever the value is in the datablse for this particular item purchased, then allow someone to select "online" or "onsite" from the dropdown if it needs to be edited? Code:
Adding From Drop Down To Database
i have a script where if you select one of the drop downs, if you have the cash in your credits it will update the database as you having one of them. HOWEVER every time it goes on it gives the player a steak house without clicking on it. anyone? Code:
2 Drop Down List Where Need Query Database
there is drop down list box linked to the database where it displays the state of a country and once the user clicks on the particular state, there purpose to be radio button displaying the locations available of that state, the location displayed in radio buttons are from the database. currently the source code i'm using is (which uses a submit button to query the database but i does not want the submit button it purpose to work using Code:
Drop Down Menu Validation Against Database
I want to do a query which selects the config.maps table.field WHERE ladder_id = $ladder_id. I can do this fine, but the next bit is slightly harder. The maps are stored as an array, separated by commas. I can explode these commas to put the maps in to an array called $maps but how can I then check that a value inputted from the drop down menu matches one of these maps? I cannot do: if ($dropdown = $maps[1] || $dropdown = $maps[2] || $dropdown = $maps[3]) { // Continue }
Horizontal Drop Down Menu From Database?
I've been searching now for months and have just given up. I've also tried to create my own menu using pre-made drop down menu software then using dreamweaver to try and get the values from a database but have had no success. I've made a CMS which allows users to greate sub menu items, and then sub-sub menu items to those submenu items. And now the final thing to do is create the drop down menu on the front end of the site. Is there a particular site or software package that anyone could point me to that is very good and allows you to create the menu's based on data in a database? So it will allow me to select which table contains the main buttons, which table contains the sub menu items, and also the sub-sub menu items.
Database Driven Drop Down Menu
I'm trying to create a database driven drop down menu. I feel I'm on to the right path but my code doesn't seem to work. I've looked around and haven't seen where I'm making the mistake. Any one care to take a look? Code:
Scriptaculous Drag Drop And Database Integeration!
I'm working on a drag and drop feature with the scriptaculous javascript library. And I was wondering if someone could help me! :( Here is some test code. Basically, when someone takes something from the left side and puts it in the right side, I just want to run a database query. I have NO IDEA how to do this because of my limited javascript experience. <script src="http://www.mytuneslive.com/_NEW/MORPH/javascripts/prototype.js" type="text/javascript"></script> <script src="http://www.mytuneslive.com/_NEW/MORPH/javascripts/effects.js?1.7.0b1" type="text/javascript"></script> <script src="http://www.mytuneslive.com/_NEW/MORPH/javascripts/scriptaculous.js" type="text/javascript"></script> <link rel="stylesheet" href="http://www.wiki.script.aculo.us/stylesheets/application.css" type="text/css" media="screen" /> <div style="height:200px;"> <div style="float:left;"> <h3>This is the first list</h3> <ul class="sortabledemo" id="firstlist" style="height:150px;width:200px;"> <li class="green" id="firstlist_firstlist1">Item 1 from first list.</li> <li class="green" id="firstlist_firstlist2">Item 2 from first list.</li> <li class="green" id="firstlist_firstlist3">Item 3 from first list.</li> </ul> </div> <div style="float:left;"> <h3>And now the second list</h3> <ul class="sortabledemo" id="secondlist" style="height:150px;width:200px;"> <li class="orange" id="secondlist_secondlist1"> <span class="handle">DRAG HERE</spanItem 1 from second list. </li> <li class="orange" id="secondlist_secondlist2"> <span class="handle">DRAG HERE</spanItem 2 from second list. </li> <li class="orange" id="secondlist_secondlist3"> <span class="handle">DRAG HERE</spanItem 3 from second list. </li> </ul> </div>
How To Grad Information From Mysql Database With Drop Down Menu's?
i am trying to do is like this website http://www.chippeduk.com/ in the search area (bottom right), where the user selects a make (all fields grab from the database), then selects a model, then they select the fine tune, then they click search this will bring up the information about the car they have chosen. How would i do this, would i use multiple tables and then link them together via an id? i am new to PHP/MySQL but very willing to learn.
Displaying Data From Mysql Database In A Drop Down List On Form.
I'm trying to do is display data from two different mysql tables from the same database in a drop down list on a html form. I have a fixtures table with the player1(userid), player2(userid), gameid, game, score1 and 2, what I want is to use the userid to get the players first name and surname from the members table (as it is a unique id), I need to do this bit before displaying it in the drop down. I think i need 2 querys to do this but when I have tried it it just echo's a blank value or the userid not the forname and surname that I want. I'm using the fetch_array function but just can't see where I'm going wrong, Code:
Getting Results From MySQL Database In PHP
I am trying to write a php page which connects to a MySQL Database which is supposed to get the results from a table within a database and display the results in a table. Below is the code that I am using:
Database Search Results
i have a paid to read website,im hosting it off my own pc..as a member you can search offers and complete them, heres where i need some help..when a member clicks submit and the offer goes to an offerdone table i need the offer to hide from them so they cant do it again,also that same offer needs to hide in the search results as well...is there somthing like onclick hide and for the results something like if duplicate hide from search or if offer found in this table hide from results... there are 2 tables one will all my offers the other with offer they have done,also this needs to apply for each member account.i am new to all this and have made it this far to have my site up and going but im looking for those extra catchs i could really use,so that a member cant repeat an offer they have already done,
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I'm dealing with a situation where I can potentially have over 3000 results to display from a database. I'm successfully paginating the results, 10 at a time, but don't want to display 300+ page links at the bottom. How can I group the pages to show, for example, the first 10 pages, followed by a "next 10" link? i.e., Page: 1 2 3 4 5 6 7 8 9 10 > >> Next 10 My current code:
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I want to know if it is possible to sort query results so that the records with a value in a certain field is displayed first and the records with the field empty thereafter.
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I'm trying to retrieve fields from my database with following mysql statement: $query = SELECT g_games.Title FROM g_collections, g_personal_reviews, g_games WHERE (g_games.ID=g_personal_reviews.G_ID AND g_collections.U_ID = '" . $data1['ID'] . "' AND g_personal_reviews.C_ID = '" . $data2['ID'] . "' ) "; $resultset = mysql_query($query); $data = mysql_fetch_array($resultset); 3 Rows in my database comply with the above statement... though I only get 1 to show up on my php page? How can I show all of them? This is the php code I'm using to write the results: while ($data= mysql_fetch_array($resultset)) { echo $data[Title]." "; } My tables are as followed: g_games g_collections g_personal_reviews > this table connects all others with G_ID ("game ID"), C_ID ("Collection_ID"), U_ID ("User ID)... users
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I have a db of tens of thousands of entries. It's not too hard to pull all the entries and build a table that displays them one at a time, but in my case, that will be a huge waste of space. What I want to do is put one record in each of the columns. I thinks it's possible to do because I have seen very elaborate thumbnail gallery scripts to do this. I just want to be able to put two or three strings of text from each record into each row, and a checkbox. Here's sample of what I mean: <table width="76%" border="1" cellspacing="1" cellpadding="1"> <tr> <td>$record1Line1<br>$record1$Line2<br> <input type="checkbox" name="$record1checkbox" value="checkbox"></td> <td>$record2Line1<br>$record2$Line2<br> <input type="checkbox" name="$record2checkbox" value="checkbox"></td> <td>$record3Line1<br>$record3$Line2<br> <input type="checkbox" name="$record3checkbox" value="checkbox"></td> <td>$record4Line1<br>$record4$Line2<br> <input type="checkbox" name="$record3checkbox" value="checkbox"></td> <td>$record5Line1<br>$record5$Line2<br> <input type="checkbox" name="$record5checkbox" value="checkbox"></td> </tr> </table> I want to pull about 200 records at a time from the db and display as described above. How can I tell php to take five records from my result set and display above, then loop through the result set until I'm done?
Filtering Results By DATE Value In Database
I have this code in my page for displaying some data from a database, sorted by date. i had posted a messed up version of this a while ago but I couldnt get it working even after a few suggestions. PHP Code:
Displaying Database And Getting Unwanted Results...
I want to display everyone in the database, but only if the 'level' field is larger than 0. Well, this works fine for most of the database... but if the last entry in the database is 0, then it prints it anyways. Is there anyway around this? I tried adding an if statement before the last bit where I echo out the table rows, but it was killing the entire table, or still showing the same results. Code:
Function To Fetch Database Results
I have a function that is supposed to fetch all the results from my table "games". There are two database records in that table. My function is supposed to display all of the records, but it only displays one. I also use a function to display a certain template. Here is the coding for both functions and my index. Code:
Counting Results From A Mysql Database
I want to know how many results throw a query in order to know if they have reached a limit.I.E. $sql = mysql_query ("select * from classifieds where user='$username'"); while ($row = mysql_fetch_array($sql)){} I want to check lets say that only 10 classifieds per 'username' can be created. If ten or more classifieds are in the database for that 'username' then show error message.
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How do I display results from a database in a particular format. For example the following text in this format including the lines in stored in the database: Quote“Hello, Welcome to the news for 11/06/07. Thanks” When I echo the results using the usual, <? echo $row_news[‘news’]; ?> it doesn’t echo the information in the database in the format it is stored for example it will display like this. Quote“Hello, Welcome to the news for 11/06/07. Thanks” How do I get it to display in the format it in displayed in the database?
Display Results From A Database On Multiple Pages
I am new to PHP and Mysql but am really determined to learn these techniques. I have some script that reads and displays results from the databse onto a page. However, I really wish to limit the number of results that are displayed to 10 per page and then have the next/privious links under the results. With these results, the results under the email field should be hyperlinked so that the person viewing the results should double click on the email link and send the person an email. How can i do that? I have to learn this technique. The code is:-
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I am loading results from mysql database and displaying on my page and then dynamically displaying only 5 results per page. On each page I have a POST button so that if user wants to POST something on form he should be able to do it whether he is on page 5 or page 1. When I first time loads the results on page 1 this POST button works fine but when I go to page 2 and click on POST it gives following error " CGI Error The specified CGI application misbehaved by not returning a complete set of HTTP headers. The headers it did return are: " and then even if I go back on page1 it starts giving me same error.
Navigating Database Results And Posting New Values
I am trying to dynamically display results that I retrieve from my database i.e. only 5 results per page and code should automatically compute how many pages it needs and create links for each page. I have also added a link at the top of my page so that if user wants to POST new data he can post it. Problem is that when I click on page 2 or 3 or any numbered page and click on POST link it gives me following error "CGI Error The specified CGI application misbehaved by not returning a complete set of HTTP headers. The headers it did return are: And it did not tell about any header etc. Code:
POST To Mysql Database And Email Results?
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Drop Down Lists Changing Depending On Other Drop Downs
I want to have a drop down system like this one. But without the radio buttons. I want to have 3 drop downs but have no idea how to go about it. I assume I will need to activate some sort of Javascript on the onChange event of the option drop downs.
Sort Results By Date/time, Limit 5 Database Entries Of Today &/or Upcoming Days
I'm designing a website for my fraternity and in the database I have entries which contain past events, todays events, and upcoming events. My problem is that I can not figure out how to sort the query results to display: 1) only past events based on the CURRENT DATE 2) only the events which fall on the CURRENT DATE or after, limiting the results to 5 entries 3) all events that fall on CURRENT DATE or afterwards Example: Today is September 15, 2007 Problem 1) Displaying all events that happened before September 15, 2007 Problem 2) Displaying only 5 results that fall on September 15, 2007 or after Problem 3) Displaying all events that are happening on or after September 15, 2007 And obviously I would like the CURRENT DATE to change depending on what date it is, currently. Ok so now I've been very redundant in my explanation of my problem here is the code and a link to what that particular page looks like. Code:
Populating A Select Box
What I want to do for members of my website is to allow them to click a button that will add their name to a list that populates a select box. I can do this on my own, but I was curious if I have to use a database in order for this to be possible.
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I have one combo box (Category) which is populated by a MySQL table using php. I am trying to use the onchange javascript to submit the value selected so that I can populate the other combo box (sub-category) based on what was previously choosen. So can anybody give me a clue on how to submit the first boxes value so that the second can see what was selected (javascript function).
Populating Dropdown List
I am wanting to know how to populate the <SELECT> dropdown menu with all existing values from the DB and also have the value associated with that id selected. This is an updating area of my admin. Any ideas? Here's the code so far. At the moment it only retrieves the value assigned to that id. PHP Code:
Dymanic Populating Checkboxes
i like to retrieve a value from DB, accordingly i like to display the checkbox, either checked or not. Now im able to retrieve the value from DB, i can show even show the checkbox, but i cant make it checked???? the Code i used to display as follows, #Require the database class require_once('../dbinfoinc.php'); #Get an array containing the resulting record $query = "SELECT * FROM event"; $result = mysql_query($query); $num=mysql_numrows($result); mysql_close(); $i=0; while ($i < $num) { if(($i % 2) == 0) { $c = '"TableDetail"' } else { $c = '"TableDetail2"' } $event_id=mysql_result($result,$i,"event_id"); $event_name=mysql_result($result,$i,"event_name"); $event_publish=mysql_result($result,$i,"publish"); // if(($event_publish) == 0) { $event_publish = 'No' } // else { $event_publish = 'YES' } $take_action = 'Delete' echo "<tr> <td Class='navText' align='center' class=$c>$event_id</td> <td class='navText' class=$c>$event_name</td> <td class='navText' align='center' class=$c><input type='checkbox' name='publish' if($event_publish==1){'CHECKED'}?></td> <td class='navText' align='center' class=$c><a href='delete_event.php?event_id=$event_id'>$take_action</a></td></tr>"; $i++; } ?>
Re-populating A Form If Errors
I am learning php. I have created a simple Web page with a form. After receing the POST request, I do some error checking. In case of errors, I would like to re-post the page, but with the current values for each field. I have organized my files like this: 1) Webpage with the form (all in html, form action points to a php file) 2) php file that receives the request.
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