Unable To Jump To Row 0 On MySQL Result
im making a login program, and i get this error.
"Warning: Unable to jump to row 0 on MySQL result index 2 in /var/www/html/login.php on line 16"
Code:
View Complete Forum Thread with Replies
Related Forum Messages:
Warning: Mysql_result() [function.mysql-result]: Unable To Jump To Row 0
i want to grab the value of a sql entry. but the thing is, sometimes this value is <NULL>. i thought, no biggy, i'll just have an if statement: if (is_numeric(mysql_result($query, 0))) { //do something } else { //do something else using mysql_result($category_id_query, 0); } however, this doe not work. i get this error: Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 3 is there a way to check to see if mysql_result() is <NULL> without throwing an error.
View Replies !
Unable To Jump To Row
I have a tariff which is atm 44 rows with 15 columns of numbers. Now i am trying to pull as individual price from the tariff based on user input: Code:
View Replies !
Unable To Jump Row 0
I am getting the following error: Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 2 in /home/content/i/n/n/**/html/tracker/connect.php on line 35 Here is my coding from line 34 on down. Code:
View Replies !
Pagination :: Unable To Jump To Row 0
I am running a script to find the "Previous" and "Next" records, the problem is when I get to record 1 I get the error below, same with when I reach the last record in the db. PHP Code: Warning: mysql_result(): Unable to jump to row 0 on MySQL result index 18 Here is the script running it, do I need to add to the query? or how can I use IF/ELSE?PHP Code: <?php $id = $_GET['id']; // Retrieve your record according to id // Display Record // Links to Previous Records $query = "SELECT id FROM jokes WHERE category = '$category' AND id < $id ORDER BY id DESC LIMIT 1" ; // For Previous $result = mysql_query($query) ; ..........
View Replies !
Warning: MySQL: Unable To Save Result Set In
I'm running PHP/MySQL/Apache on my home system on Windows for testing purposes. As a run a specific query, I get this error: Warning: MySQL: Unable to save result set in C:apachehtdocsfinal.php on line 129 Warning: Supplied argument is not a valid MySQL result resource in C:apachehtdocsfinal.php on line 132 I know the second result means the specified query cannot be found, and I'm guessing for some reason MySQL cannot save the query. Here is my code. PHP Code:
View Replies !
Warning: MySQL: Unable To Save Result Set
I'm getting this error: Warning: MySQL: Unable to save result set in /www/hosts/wwwroot/mainPage.php on line 11 Here is what line 11 looks like: $articles = mysql_query("select entryID from articles where siteID like 'senior'");
View Replies !
PHP/MySql Jump Script
I have also read in a affiliate forum (quote below)that putting &afsrc=1 at the end of a link will help click through, how might this also be implemented? "Recently, I put the &afsrc=1 on all my links across my largest sites. Damned if sales have not increased by 10-20% since then. (it has been about 2 weeks at about 13000 clicks a day)" <?php $link = mysql_connect ("localhost", "user", "pass"); mysql_select_db ("datebase"); $result = mysql_query ("SELECT link FROM id WHERE number=$locn"); $row = mysql_fetch_array($result); header("Location: " . $row["link"]); mysql_close ($link); ?>
View Replies !
Unable To Start MySQL
I have just installed, PHP, Apache, on my WinXP Pro machine. I then went to install MySQL, which appeared to go well. I went into MySQL folder and opened up winmysqladmin, and clicked on Start. the response I got was unable to start try again in 30 seconds. Don;t know if this has anything to do with it, but in the books and help pages I have got, they all mention a log on page, to enter username and password, I have not seen this.
View Replies !
Unable To Connect To MySQL
I've finished coding php, run it OK with my localhost and uploaded to my Apache server with Linux OS. However, I'm seemingly unable to connect to mysql at the server side... and received the following message: Could not connect to database: Can't connect to local MySQL server through socket '/tmp/mysql.sock' (111) What's this message abt? I tried to get into MySQL without php on the server.. and received the same message at the console... how come?
View Replies !
Unable To Detect MySQL Support
"The installer was unable to detect MySQL support in PHP. Please ask your host to ensure that PHP was compiled with MySQL, or that the proper extension is being loaded." This is the error i am getting while installing a free software - Simple machines forum. What can be done to install the software. I am trying to do it on windows platform.
View Replies !
PHP:Unable To Post Variables From Html Form To Mysql Database
Plz dont treat this as another newbie query , i did my homework but still getting nowhere :( :( :( Trying to learn PHP on Fedora core 1 (PHP 4.3,MySQL,HTTPD).Unable to post data from html form to php file(connecting to mysql database and inserting into a table) . This seemingly simple problem is making me go mad !!! I googled extensively and already tried the following I enabled " register_globals = On " .... in /etc/php.ini and also took used _$POST syntax. wot could be the problem .... is it the case that /etc/php.ini file is not being read .... how do i know whether this is the default location for php.ini? Where can i find the PHPRC environment variable (the variable responsible for the php.ini default location) Is the problem something else .....? Following is the code ... form.html --------- <HTML> <HEAD> <TITLE> email entry form </TITLE> </HEAD> <BODY> <P>plz fil the form </P> <FORM METHOD="POST" ACTON="form-handler.php"> NAME: <BR> <INPUT TYPE=TEXT NAME="givename" SIZE=25> <BR> AGE: <br> <INPUT TYPE=TEXT NAME="givenaddress" SIZE=25> <BR> <INPUT TYPE=SUBMIT> </FORM> </BODY> </HTML> form-handler.php ----------------- <HTML> <HEAD> <TITLE> EMAIL FORM HANDLER </TITLE> </HEAD> <BODY> <? mysql_connect("localhost","root") or die("Failure on db connection"); mysql_select_db("mydb");
View Replies !
Last Row Of Result MySQL
I am making a wrapper class for MySQL functions and I want to be able to return the last row from a query. here is the function in my class PHP Code: function getLastRow(){ $this->lastRow = mysql_data_seek($this->result,$this->numRows-1) return $this->lastRow; }//end getLastRow I am using mysql_data_seek and pointing it to the last row, but it is not returning anything.
View Replies !
MySQL Result Resource
I want to build a function that, depending on the input, will either return the results from a MySQL query or do something else and return one of several messages. My problem is checking the return value to see which is returned. Is there a php function that checks a variable to see if it is a valid MySQL result resource? I looked through the manual and couldn't find anything.
View Replies !
Returning 1 Result From MySQL
I know the usual way of receiving results from MySQL, useing a while loop e.g. $sql = "SELECT * FROM users WHERE email='joe@hotmail.com'"; $result = mysql_query($sql,$connection); while ($row = mysql_fetch_array($result)){ echo $row["firstname"]; echo $row["lastname"]; echo $row["email"];} BUT, if I know they only 1 result is going to be returned, is there a way to do this without using a while loop? So if I only wanted to get only the firstname of the person with email joe@hotmail.com. can I do this without a while loop or do I have to use one?
View Replies !
Total Result Php Mysql
PHP Mysql limit the result to 5 I can display the 5 results using a do & while which it ok. I want to list the total or all the results without using a do or while this is so I can combine the results and remove duplicate words from the total results I.E. Do & While row result 1 this is the top 2 this is the bottom 3 this is the middle 4 this is the end 5 this is the finish This is what I want to do: result this is the top this is the bottom this is the middle this is the end this is the finish modified result:: this is the top bottom middle end finish
View Replies !
Invalid MySQL Result
I keep getting the following two errors: Warning: Supplied argument is not a valid MySQL result resource in /home/xxx/public_html/articlepro/newarticle.php on line 75 Warning: Supplied argument is not a valid MySQL result resource in /home/xxx/public_html/articlepro/newarticle.php on line 84 Here's the code which I think is causing the problem. PHP Code:
View Replies !
MySQL (DB) Result Sorting
I'm doing is taking results (fetching rows) from my mySQL database. But what I want to do is sort the results according to last name. So basically the table structure looks something like this: id (primary key) fullname A example rows could be: 1 John Smith 2 Mary Jane Now what I want to do is sort the results in PHP based on last name. I'm using the Pear DB for connecting to my mySQL database. Based on the example rows I gave, Mary Jane would go first, and then John Smith. How would I go about doing this? I was beginning to split the result based on spaces, and the last word to explode, but how would I truly do this?
View Replies !
Modifying MySQL Result - But Not The DB
How can I modify a MySQL data set returned by mysql_query? I am basically doing a while loop over the rows, changing one field, and then doing a mysql_data_seek back to the beginning of the set before returning it. However, later calls to mysql_fetch_assoc still return the orginal data, meaning that (I assume) I am changing a copy of the data rather than the result set itself. Taking the reference (i.e. $record =& mysql_fetch_assoc($result);) did not help. I have read a large quantity of the posts in the php.net site, googled, and read my PHP books - but have come up short on this one. Yes, I know it's better to simply modify the DB. However, I'm working with an existing code base which is very complicated, and want to cherry pick one very well tested feature by making a change in the result set only under special circumstances.
View Replies !
Mysql Result Into Array?
I have a shopping cart script sending info to a processor. I need to send the qty's and item name's in some sort of string via a single variable to the process form. i.e. (3) Hipster Turnips, (6) Butter Milk Baby Brains, (2) Super Freaks Code:
View Replies !
Result From Mysql Into Foreach?
when using mysql_fetch_array($result) I usually toss that into a while loop like so while($row = mysql_fetch_array($result)){ ... } however I was wondering if there was a way to put it in a foreach statement like so foreach(mysql_fetch_array($result) as $row){ ... } Is that legit? Will this work?
View Replies !
Linking To A MySQL Result
I had it the other day but now altered the code for my real database and it stopped working. Here is the first page which lists the 10 most recent news stories: <?php require ('get_connected.php'); $sql = mysql_query ("SELECT title, story_id FROM news ORDER BY story_id DESC LIMIT 0, 10"); while ($row = mysql_fetch_assoc($sql)){ echo "<a href='view.php?id={$row['story_id']}'>" . ucwords(strtolower($row['title'])) . "</a><br>"; } ?> And here is the page that should be displaying the news story based on story_id selected by the user.....
View Replies !
Reverse Mysql-result
Is there any way, to reverse the result of a mysql query? Explanation: If i have eg. 20 records, all with their own id of course, select 5 with limit, and order them DESC by ID, it should give me: 20, 19, 18, 17, 16. That's all ok. But now i want 16 to appear first, 17 second, etc.
View Replies !
Mysql Result Set As An Array
the question is, is there a way of assigning each value returned from the sql query and assinging them to a single array. for example i have a a query like so; $table = "sales_table";         $query = "select treatments from $table where date between  ད/08/2006' and ཕ/08/2006'"; $result = mysql_query($query); because the values stored in the treatments field are imploded as a single string when the user submits the form the value looks like; 1, 22, 33. Code:
View Replies !
Not A Valid MySQL Result
I have a login system that worked perfectly on my test server but once I moved it to the LIVE server I now get an error when I try to get a list of all the people signed up at the login system. Code:
View Replies !
Characters From A Mysql Result
How would I go about limiting the number of characters from a mysql query to only like 20 characters and if its more than that then it will add "..." to the end of the result.
View Replies !
Mysql Query Result
I have successfully selected 4 rows using a mysql SELECT query. However, I would like to random pick a row from this result so that every page refresh will yield something different.
View Replies !
Mysql Table Result
I have a product page which grabs a product, lets say the ID = 1. At the bottom of the page it grabs the 2 latest products created by that user. I would like it to grab the next 2 products in the database instead of the latest. So, Product 1 Other Products Product 5 and Product 6.
View Replies !
Setting A Php/mysql Result
I can't seem to figure out on my own using tutorials or documentations so I was hoping someone here could help me out. Here is my problem. I am using PHP and MySQL to do this script. I want to have a table that contains only numbers for data and then add them all up as a result to display and that I have accomplished with no problem and then I want to display a percentage of that number to the public however that also I have accomplished with no problem using round(). I want it to display 1-500 no matter what the number is or the decimal it may have is provided it doesnt exceed 500 but once it does I want it to display 500 as the limit so even if it is say 674 I want it to only display 500. The sql I am using for the adding is SELECT SUM which works just fine and I am using round( $total_number *.15, 2) to display the percentage number of 15%.
View Replies !
MySQL Result As Array
$db= new MySQL($host,$dbUser,$dbPass,$dbName); $sql=("SELECT * FROM sticker_files WHERE sticker_cat_id=Ǝ'"); $catdirs = $db->query($sql); $dirs=$catdirs->fetch() and I need a for each loop (i think) to insert that data below: $params = array( 'mode' => 'Jumping', 'perPage' => 3, 'delta' => 2, 'itemData' => array(****RESULTS HERE AS COMMA SEPARATED string) );
View Replies !
Mysql Result To Array
I have this query, it gives the suspected information in myphpadmin: select listid, count(*) from table1 group by listid How do I put this the result into an array in php? I have problems with the count(*) field
View Replies !
MySQL Result Object
I require two copies of a mysql result object. I am simply doing: $result = mysql_query($sql) or die(mysql_error()); $temp_result = $result; But when I call: while($row_temp = mysql_fetch_object($temp_result)) It seems to unset the $result object as well.
View Replies !
Result Mysql As Variable
I have an select on which the user choose property, arrival and departure dates for calculating the price, works perfect. But I need to put in some sort of error if the person choose dates higher than end date in the database, What I want to do is this, but don´t work of course: sep_fin is an datecolumn in the database if ($salida>"sep_fin") { echo "¡ There are no prices yet!"; } elseif ($row = mysql_fetch_array($result)){
View Replies !
Random Mysql Result
ive been trying to make a random mysql result grabber which i have done. But what id like to do is have it stick around for awhile. So i was wondering how i can give it a specific time and after that time it will grab another result.
View Replies !
Array With MySQL Result
//MAIN FLOW $counter = 0; $sql[$counter] = "SELECT * FROM category WHERE parent = 0 ORDER BY parent"; $result[$counter] = mysql_query($sql) or die(display_and_log_error("SQL ERROR: ".__LINE__)); while($myrow[$counter] = mysql_fetch_assoc($result[$counter])){ echo $myrow[$counter][category_name]; ...
View Replies !
Using Jump Menu
I am using a jump menu to select multiple drop down boxes in a form which are related to the choice you make in the first drop down box. The problem is that the page reloads and everything before this tag is cleared. I want to be able to repost the data. I know that I need to use sessions in my code but I dont know exactly how to use them for this case. Code:
View Replies !
Jump Frame?
I want to make a jump frame (i think is the name), when people click in an image or video, appears a frame in the top (or left), but i want to put an id for each video and image. There are a simple and better way, but i don´t know how to do, where we can put only the link and it redirects with the frame in php (out.php) like this one (this is an adult link, not spam, only is an example what i want): Code:
View Replies !
JUMP Out Of The Script
"mysql_connect establishes a connection for the duration of the script that access the db. Once the script has finished executing it closes the connection. The only time you need to close the connection manually is if you jump out of the script for any reason. " http://www.webmasterworld.com/forum88/903.htm What does he mean by jump out? I used header redirection, is this "jump out", but after header redirection, I always added exit;, does it mean that I need to close the connection.
View Replies !
Jump Menu
<select name="menu1" onChange="MM_jumpMenu('parent',this,0)"> <option value="www.site1.com">google</option> <option value="www.site2.com">metu</option> <option value="www.site3.com">mynet</option> <option value="www.site4.com">asa</option> </select> above code ; i want to get the selected value and post as named $url into a php code. How can i call the selected value?
View Replies !
Using MYSQL Result To Send Mail
I have wrote a program to load a number of email address from the MYSQL database. And I have use "$mail = mysql_query("select email from user;",$link_ID); Now, I want to use mail() to send email to those e-mail account I got. What should I do for this?
View Replies !
Date Formatting From Mysql Result
always in the past I have done my date formatting from the query like so: DATE_FORMAT(dateField, '%W, %b %e, %Y') AS realDate I have tried formatting the straight date result with PHP like so: having fetched results with mysql_fetch_array, $newDate=$result["dateField"]; $formattedDate=date($newDate, 'm/d/y'); This doesn't work and simply returns the unformatted date as it is stored in the database.
View Replies !
Invalid MySql Result Resource
I have written a php script to search a MYSQL database and with the line: $result = mysql_query ("SELECT * FROM table1 WHERE first_name LIKE '$first_name%' AND last_name LIKE '$last_name%' " ); The next line, if ($row = mysql_fetch_array($result)) { , gives me the error message "supplied argument is not a valid MySql result resource"
View Replies !
Display Of Mysql Table Result...
I have a mysql table of articles with fields: - recordID - department - articleTitle - articleText Using PHP, I'm attempting to get the results of the table to display as follows: Department 1 - articleTitle 1 - articleTitle 2 - etc.... Department 2 - articleTitle 1 - articleTitle 2 - etc.. I figure I should use a while loop but can't figure out how to exit/reset the loop when a new 'department' is encountered in the $result.
View Replies !
Replacing Array With Mysql Result
I wanted to repalce the following line of code $data = array(40,21,17,14,23); with for($i=0;$i<$numrows;$i++) { //print(mysql_result($result,$i,2));print("<br>"); array_push($data,mysql_result($result,$i,2));} where mysql_result($result,$i,2) is the value and when i print it displays the values. But $data array is the Y axis value for drawing a chart. Here I wanted to replace the hard coded value with values from mysql but the second code does not function. Does anybody have idea how can I replace the $data array or what may be the problem with my coding The first one works but the second one does not work but in both cases it does not display any error.
View Replies !
|
TRACKER
