Selected Value In Dropdown Dependent On Database Value
i need a way to make the selected value of a drop down dependend on a result from a database query. PHP Code:
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Dependent Dropdown
I have set up a MySQL database with 5 tables that are relevent to this question. The first table is for entering records, the second has only 2 foreign keys that join the first and the third together, the third has a primary key for numerical value and a name column, that is used for the first dynamic dropdown list, the forth table, like the second has only 2 foreign keys to join the third table and the fith table and finally the fith table, like the third table has a primary key for a numerical value and a name column for the planned dropdown menu. How I can get the first dropdown menu to work, how do I get the second one to run off it and the form or forms need to submit to 3 tables, how do I do that?
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Php Dropdown - Selected?
i'm using this to pull information from a database and display it in a dropdown list. The problem is I am using the name code when I goto the "Update" page, but i'm not sure how to add the "selected" field, that way what ever I saved in the database will load. Example: I goto add and it add's the "id" of 7 to the field, when I load the page again it auto's back to 1 instead of clicking in-to 7 as thats whats already there... = ) Code:
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Remember Selected Value DropDown
I am attempting to have a drop down menu that displays url from 1 table called links the same table also contains a name for the url. i.e. google.com = name and http://www.google.com = url The value of just the url from links is stored in a separate table called users in a field named search after form is submitted by user. Everything Posts correctly except when go back to edit the record and submit the form the drop down box does not retain the value of what had been previously selected although it is the correct record until form is updated when whatever value is currently in the drop down list will update the record. I know the code is doing exactly what it is told to do, because I have no functionality in it to recall the value from the field search in the users table. I don't know if it is something I would have to add to the array or option value. I can get it to work if it is all hard coded, but I am attempting not to do this. I am pretty new to php mysql and any help is greatly appreciated. This is the code I am working with that came mostly from a tutorial. PHP Code:
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Dropdown Menu <selected Name="...
I have a dropdown menu like this one: <select name="country" tabindex="6" id="select_country"> <option value="">- Country List -</option> <?php $query = "SELECT * FROM country ORDER BY country ASC"; $result = mysql_query($query) or die('Error, query failed'); while ($row = mysql_fetch_array ($result)) { $country = $row['country']; ?> <option value="<?php echo $country ?>"><?php echo $country ?></option> After I have selected one country and I hit the submit button it go back to the default value in the box, what I want is to show the selected value in box until I selected another value. If the default value is Australia when I start this page, I then select USA and hit the submit button it reset to the default value "Australia", but I want it to show USA as long I not select anything else. I have tried to use session in selected="<?php echo $_SESSION['country']; ?>" but it didnt work.
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Print All Selected Items From Dropdown??
I have a program that allows users to select a 'New Graduate' membership, and for each one selected, they get a free tshirt. I've got that working using a 'is_new_grad' function that I created. So, if they select 2, they see 2 dropdowns to select the sizes of them. It then sends an email w/ the membership info and the shirt sizes selected. However, when the email is being sent, it's only returning the size of the LAST item selected in from the drop-downs. Example: if I sign up for four of these memberships and select four different tshirt sizes (1 Medium, 1 Lg, 1 XL, and 1 XXL), the info in the email is all XXL: Code:
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Dropdown List Selected By Default?
Does any one know how to have an item in the dropdown list selected by default? $city_field = HTML_QuickForm::createElement('autocomplete', 'city', 'City', $cities); $cities is an array of city names and I would like, say 'Los Angeles' selected by default.
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How To Pass The Value Of Selected Dropdown Menu To Another File?
there is a php-myql script that list the mysql databases in the drop-down menu, use wil select one and the submit button should pass the value or variable $db_name to the test.php, this script list the databases in the drop-down menu but when i select one and then click on submit cannot pass the db_name to the test.php:
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Must Database Table Be Called Up For Dependent List Box Form?
I want to produce a form which will post selections named size and drive to to a file blindsize.php. Looking around this forum and the web, it seems that the only way to do this is to set up a database table (I would set up one in MySQL) and call up that with javacript. If setting up a MySQL table is the answer, then I would probably use this script. But as I am also new to javascript, I would now have a further learning curve with that, though at first glance the script tutorial looks very good. But is setting up a database table the only way for a php user? When I first started looking around, I had though the way might be a series of elseif lines.
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No Database Selected
Im running twice the same website: - one for the public - one for private testing Everything is moreless the same, except some config variables in the php and the replacement of css directories. However on the test server im getting a no database selected: mysql_query('sqlhere') doesnt work mysql_query('sqlhere',$conn) does work So somehow on my test server it is not taking the only open connection, but wants it specified. Code:
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Database Selected Error
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1 Record In Database But 2 Records Were Selected?
i'm retrieving a record from database. below the record, i have a yes and no button, where i can let user click and then count the number of yes and no. everything works fine, but when i start to click the yes or no button, my record ,becomes two records. i only have one record in my database, but two same record was displayed. this is the code: PHP Code:
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Populating A Dropdown From A Database
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Add Data To A Database Using Dropdown Menus?
lets say i had a database with a table called admin hacks and another table called user hacks,and both tables had feilds called name and id now what i want to be able to do is create a submit form with a drop down menu saying the table names mentiond above, also the form will have a field called name. now lets say a user slected admin hacks from the drop down menu and enterd jack in the name, how would i get the data in that submitted form to be inserted into the correct database table, so jack should now be in the table called admin hacks also lets say someone selected user hacks and entered jon how would i do the same for that?
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I be able to insert the date into my database if i have 3 dropdown boxes wherein it contain the month,day & year. i know the basic of inserting data into my database but with this 3 dropdown box i don't have any idea. after submitting, the data selected on these 3 boxes shall be in one field in my database with fieldname Date.
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If Else Dependent On Variable In Sql
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Dependent Drop Downs?
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Dependent Drop Down Lists
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Dynamic Dependent Drop Down Boxes
I need to build a dynamic page with 2 drop down boxes on it where the contents of the second drop down box are dependent upon the selection made in the first drop down box (the first box contains districts/regions and the second box contains towns/cities). The customer wants this to happen on a single screen but can this be done with PHP?
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Sending To A Second Email Dependent On Checkbox Value
I have searched this board and others, I have looked through my books and I can not figure this out. I have tried at least a dozen variations and none of them seem to work. I have a form that is on an html page. When the user submits it, it sends an email to person 'A', but if they select a checkbox then a copy has to also go to person 'B', but only if the box is checked. I figured I could just put the php in the form to create the hidden input, not getting it. Code:
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Multiple Selected="selected" In A List.
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Selected Value
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$selected
I'm trying to generate a select box through php. the problem comes when I try to dynamically add in the 'selected' attribute to one of the options. My script is basically the following: Code: foreach($values as $key => $value) { if ($key == $checkVar) { $selected = ' SELECTED' } else { $selected = '' } $options .= '<option value="'.$key.'"'.$selected.'>'.$value.'</option>' } print '<select>'.$options.'</select>'
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Know Out Of All Checkboxes How Many R Selected And Which Are They?
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Selected In A Drop Down
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Resize Only Selected Images
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Radio Button Selected
What I wish to do is have a radio button pre-selected depending on what is pulled from the database. For example if I had 2 radio buttons: o Yes o No One has the value 1 and the other has the value 0 If 1 is selected from the database then Yes is pre-selected, else 0 is pre-selected. How would I do this?
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Setting Selected Option Using PHP
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Jump Menu Selected Value
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Collecting Selected Options
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