Mysql_fetch_array Error - Warning: 4 Is Not A MySQL Result Index
I have written a function to bulid a table of url links stored in MySQL. In my database I have two tables: tblLinkTypes and tblLinks. My Output should look like this:
Link Type1
Link Title2 URL2
Link Type2
Link Title1 URL1
However I get this error when the code gets to the bottom (see //errors here).
Warning: 4 is not a MySQL result index in /home/httpd/inc/dbWhiteWater.inc on line 275
Code:-
View Complete Forum Thread with Replies
Related Forum Messages:
Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid MySQL Result
Here is my script: <?php $dbh=mysql_connect("localhost", "my_username", "my_password") or die ('No can do!'); mysql_select_db ("my_databasename"); $query = "SELECT * FROM contacts"; while($i = mysql_fetch_array($query)){ echo $i[first]; echo $i[last]; echo $i[email]; } ?> Get this error message: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home2/my_username/my_domain/my_databasename/test.php on line 6.
View Replies !
Warning Mysql_fetch_array(); Supplied Argument Is Not A Valid MySQL Result Resource?
i am getting the following error with this bit of code? Warning mysql_fetch_array(); supplied argument is not a valid MySQL result resource? Code: <?php /* Connect to database */ $extra = 'blah' $size= Ƈ' $paper= 'matte' $db= mysql_connect("localhost", "", ""); mysql_select_db("", $db); $sql="select type_id FROM type WHERE extra=".$extra." AND size=".$size. " AND paper=".$paper ; $type_id=mysql_query($sql); ............
View Replies !
Error Message - 0 Is Not A MySQL Result Index
Could anyone explain what this might mean? Warning: 0 is not a MySQL result index I have a simple form set up where a user types in a keyword and then a page is displayed based on the keyword. See below: HTML page: <html> <head> <title></title> </head> <body> <form action="submitform.php3" method="GET"> <p><strong><font face="verdana, arial, ms sans serif" size="1">id#</font>:</strong> <input type="text" name="keyword" size="15" maxlength="25"> <input type="submit" value="Go!"> </p> </form> </body> </html> submitform.php3 page: <?php mysql_connect ("localhost", "username", "password"); mysql_select_db (dbname); $result=mysql_query("SELECT $url FROM table WHERE keyword LIKE '%$keyword%'"); list($url)=mysql_fetch_row($result); ?> The concept is really simple. A user types in an keyword and that page is displayed. There is a database set up with a table that has two columns. One is "keyword" and the other is "url" For example row 1 the keyword is abc, and the url is http://www.abc.com.
View Replies !
Mysql_fetch_array Result Type Error
I'm trying to run mysql_fetch_array with a variable in place of the optional result type constant, like so mysql_fetch_array($result, $rtype) where, obviously, $rtype is one of the three valid values for the result type. This throws an error: "mysql_fetch_array() [function.mysql-fetch-array]: The result type should be either MYSQL_NUM, MYSQL_ASSOC or MYSQL_BOTH." I'm not quite sure what the issue here is, other than that I'm replacing a constant with a variable. I'm looking for a solution that will allow me to (variably) be able to dictate this value.
View Replies !
About PHP::: Error/Problem PHP Warning: Mysql_fetch_array():
About PHP::: Error/Problem PHP Warning: mysql_fetch_array(): ================================================== In the past i use my code/script work fine with Mandrake 9.1 - 10.1 ( + apache ,mysql ) Now i move to redhat EL4 with the same code/script I got error can't login to my code. and found error on /var/log/httpd/error_log like this PHP Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /var/www/html/website_in_server/showdata.php or PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /var/www/html/website_in_server/showdata.php I think problem source from php.ini or httpd.conf .
View Replies !
Checkbox Value - Result In An Undefined Index Error Message
I have some checkboxes which if you leave unchecked, result in an Undefined index error message. How can I get round this so there is some kind of default value that is entered apart from the checked or unchecked value: PHP Code: <td>VDE (Volunteer Development England) Quality Mark</td> Â Â Â Â Â Â Â Â <td> </td> Â Â Â Â Â Â Â Â <td><input type="checkbox" name="list_quality_sys_29" value="Heard of this"></td> Â Â Â Â Â Â Â Â <td><input type="checkbox" name="list_quality_sys_30" value="Relates to organisation"></td>
View Replies !
Mysql_fetch_array :: Supplied Argument Is Not A Valid MySQL Result Resource
I've got a problem with mysql_fetch_array function. I have two function: function db_query($dbh,$qw) { db_test($dbh); $result = mysql_query($qw); return $result; } function db_insert($dbh,$table,$what,$info) { db_test_table($dbh,$table); $id=0; $qw="insert into $table $what values $info"; $result = db_query($dbh, $qw); return $result; } Then I have the following piece of code: $result = db_insert($dbh, $table1, $what1, $info1); $row = mysql_fetch_array($result); The db_insert call workes fine, 'cause a new record is inserted into the table. The problem is that when I call mysql_fetch_array I got the following error: "Supplied argument is not a valid MySQL result resource".
View Replies !
Mysql_fetch_array(): Supplied Argument Is Not A Valid Mysql Result Resource
i recieve this error from this part of the code: mysql_fetch_array(): supplied argument is not a valid mysql result resource it's returning a long text from the database which is about 200 characters long. <?php if ($_SESSION['is_town'] = 1){ $query = "SELECT usemap FROM town WHERE town_id = " . $_SESSION['townID'][0]; $db_result = mysql_query($query); $db_usemap = mysql_fetch_array($db_result); echo $db_usemap;} ?>
View Replies !
Nested MySQL Queries Create Issue With Validity Of Result Index?
I am iterating through a result set to generate a second set of queries but no matter what I do I get the error Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource even though if I echo the query to the browser and cut and paste it into the command line I get valid results. Do I have to store the first result in an array before doing the second set? while ($i < mysql_num_rows($result)) { $PIDs[$id]['ClientName'] = $client; $PIDs[$id]['ProjectName'] = mysql_result($result, $i, 'Name'); $PIDs[$id]['PID'] = mysql_result($result, $i, 'PID'); $thisPID = $PIDs[$id]['PID']; $PIDresult = ("SELECT * from hours WHERE PID = '$thisPID'") or die(mysql_error()); $j=0; while ($j < mysql_num_rows($PIDresult)) { $PIDs[$id]['PID']['Comments'] = mysql_result($PIDresult, $j, 'Comments'); $j++; } $i++; } }
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: $articles = mysql_query("select entryID from articles where siteID like 'senior'");
View Replies !
Warning: Supplied Argument Is Not A Valid MySQL Result Resource
<? Warning: Supplied argument is not a valid MySQL result resource if (mysql_num_rows($result) >0 ) !Didnt have this problem before session_start(); if ($username && $password) { // if the user has just tried to log in $db_conn = mysql_connect("localhost", "*****", "*******"); mysql_select_db("micro_db.adf", $db_conn); $query = "select * from users " ."where name='$username' " ." and pass=password('$password')"; $result = mysql_query($query, $db_conn); if (mysql_num_rows($result) >0 ) { // if they are in the database register the username $valid_user = $username; session_register("valid_user"); } } ?>
View Replies !
Warning: Mysql_result() [function.mysql-result]: Unable To Jump To Row 0
i want to grab the value of a sql entry. but the thing is, sometimes this value is <NULL>. i thought, no biggy, i'll just have an if statement: if (is_numeric(mysql_result($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 !
Warning: Smarty Error: Unable To Read Resource: "index.tpl" In
I am using "Smarty 2.1 Version" in php. It is giving following error Warning: Smarty error: unable to read resource: "index.tpl" in /home/httpd/tsils/parag/tradesite/Smarty.class.php on line 1095 the directory which is containing templates i.e. templates and templates_c is having 777 permission. Also tried with full absolute path in the Smarty.class.php. like var $compile_dir = '/home/httpd/tsils/parag/tradesite/templates_c'
View Replies !
LEFT JOIN - Get The "Warning: Mysql_fetch_array(): Supplied Agrument Is Not A Valid MySQL Resource".
I am trying to merge results from two tables, but this is something I haven't done so far, so manybe my mistake will look stupid to Pro's. I get the "Warning: mysql_fetch_array(): supplied agrument is not a valid MySQL resource". Here is my code: $result = mysql_query("SELECT * FROM az_mustseethis LEFT JOIN az_articles ON az_mustseethis.art_id = az_articles.art_id WHERE az_articles.category = 'Must See This' ORDER BY art_id DESC LIMIT $from, $max_results",$connect); Instead of adding the latest entry in the "az_mustseethis" table, I want to take it from the "az_articles" table, where it has already been added.
View Replies !
Warning: Mysql_num_rows(): Supplied Argument Is Not A Valid MySQL Result Resource In *** On Line 9
I keep getting an error result saying: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in *** on line 9 Warning: Cannot modify header information - headers already sent by (output started at ***:9) in **** on line 12 I'm pretty new to PHP so I'm thinking its a problem with syntax. Any help is appreciated. Code is below. <?PHP $ni = trim($_POST['ni']); require 'db_connect.php' $sql = "SELECT * FROM $db_table WHERE ni = '$ni'"; $query = "mysql_query($sql) or die (mysql_error())"; if (mysql_num_rows($query) < 1) { require 'db_close.php' header("location: DESIRED LOCATION"); } else{ require 'db_close.php' header("Location: EXIT "); } ?>
View Replies !
Warning: Mysql_num_rows(): Supplied Argument Is Not A Valid MySQL Result Resource In /mounted-storage/
my code: $username = mysql_real_escape_string($username); $result= mysql_query("SELECT * from `admin_login WHERE username=".$username); if (mysql_num_rows($result) != 0) { echo ("That Username is already being used. Please try a different username."); } I know that there is a row that has the same username as $username, but i keep getting an error like: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /mounted-storage/home23a/sub001/sc21473-GRUR/www/inphp/members/register.php on line 56 Line 56 being: if (mysql_num_rows($result) != 0)
View Replies !
Mysql_fetch_array Index Not Working
I'm doing this: <?php $var = $row['title_en']; ?> that works perfectly and fetches from the fieldname named: title_en now I do this: <?php $lang = 'en'; $var = $row['title_'.$lang]; ?> it doesnt work so i try this: <?php $var = $row["title_$lang"]; ?>still doesn't work my goal is to call the fieldname: title_en using the variable $lang which equals: en p.s. I have also echoed out $lang, and it does indeed equal: en how do I do it?
View Replies !
Issues With Using Mysql_fetch_array($result)
I keep having serious issues with using mysql_fetch_array($result) and I don't know why. I have used it many times previously without issue. Here is the code I am using (I have added in a few items while trying to debug it). Code:
View Replies !
Inaccurate Mysql_fetch_array Result
When I use this code: while($row=mysql_fetch_array($result)){ it DOES work, BUT it always gives me ONE LESS out of the array than it should. ex) If I have 5 items in the array, it'll only print out 4, etc when I use the echo commands. PS. I fogot to thank the puerson who helped me last time when I posted here. :headsmack: His suggestion did work! Now I can't find it.
View Replies !
COUNT() - Warning: Mysql_fetch_array():
PHP Code: $url_count = mysql_query("SELECT COUNT(*) FROM unique_views WHERE page_id = $myrow[page_id]"); $urlrow = mysql_fetch_array($url_count); I get this error: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in.
View Replies !
MySQL Free Result Error
I'm getting a free result error when trying to load my page. It loads, but this is as far as it goes. Any idea what's the cause or how to fix it? I'm not familiar with such an error. Warning: mysql_free_result(): supplied argument is not a valid MySQL result resource in department.php on line 226. Here's my page code. It's supposed to just be a database page that pulls information from the category called by the URL. Code:
View Replies !
Function To Test If Mysql_fetch_array($result) Pointing Last Record?
I try to perform multiple insert statment, but the problem is in the below codes, the last sql concatenation ends with ",". I wonder if there is any php function to test if the $result set contains the last record. ------------------ //Get topic id of vote_topic from last insert record $sql_topic_id = "SELECT topic_id FROM vote_topic ORDER BY topic_id DESC LIMIT 1"; $result = mysql_query($sql_topic_id,$conn); $row = mysql_fetch_array($result); //Get all student info from stuinfo $sql_strn = "SELECT strn from stuinfo"; $result1 = mysql_query($sql_strn,$conn); //Dump all student strn into vote_status. //Multiple insert statement? How? $sql2 = "INSERT INTO vote_status (strn,topic_id) VALUES "; while ($row1 = mysql_fetch_array($result1)){ $sql2 .= "VALUES ('".$row1[strn]."','".$row[topic_id]."'),";} mysql_query($sql2,$conn); ---------------------
View Replies !
An Extra Last Empty Field In An 'mysql_fetch_array' Result Array?
I seem to get an extra empty field in every 'mysql_fetch_array' command I issue. For example: I have a simple table 'tblName': ID Name 1 Jane 2 Joe 2 Doe The following code: $oCursor = mysql_query("SELECT ID from tblName WHERE Name='Jane'"); if (!$oCursor) { $bGo = false; } else { $aRow = mysql_fetch_array($oCursor); } results in: count($aRow) = 2; $aRow[0] = 1; $aRow[1] = '' Am I missing something, doing something wrong, a wrong PHP setting?
View Replies !
Warning: Mysql_fetch_array(): Supplied Argument Is Not A Valid
When I run this script It says Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in blah blah. <? session_start(); include "db.php"; $username = $_SESSION['username']; $result = mysql_query("SELECT * FROM users WHERE to='$username'"); while ($row = mysql_fetch_array($result)) { ?> From: <? $row['from'] ?><br> Subject: <? $row['subject'] ?><br> Message: <? $row['message'] ?><br> <? } ?>
View Replies !
Mysql Error - Warning: Mysql_numrows():
Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in e:phpdevwwwdatabasemain.php on line 7 Filename Type Size Description my code for that is as: $sql = "SELECT * FROM tbl_Files "; $sql .= "ORDER BY filename ASC"; $result = mysql_query($sql, $db); $num = mysql_numrows($result); for ($i = 0; $i < $num; $i++) { $data = mysql_fetch_object($result);
View Replies !
Mysql - Error Warning Cant Connect To Localhost.
simple problem i cant seem to sort out. i can login to mysql account through mydb studio windows program but i try to login through php script // Database Connection $connect = mysql_connect("localhost", "root", "admin") or die(mysql_error()); mysql_select_db("databasename") or die (mysql_error()); i get error warning cant connect to localhost.
View Replies !
Mysql_free_result Return Error Saying Not Valid Mysql Result Resource
im trying to do a query which takes rows of large amounts of text data and displays them one after another. the size of the fields is causing the page to be massive in size. so to try and reduce this im trying to use mysql_free_result to clear the contents from memory. however its returning an error saying not valid mysql data when im certain that it is. is this because im using a while loop like this...? ($row = mysql_fetch_result($result)) { mysql_free_result($result); }
View Replies !
Warning: Odbc_connect(): SQL Error: [MySQL][ODBC 3.51 Driver]Unknown
I'm trying to connect to an Access database via a system DNS called PromoDB which has a username of root but no password, and I am currently getting the following error: Quote: Warning: odbc_connect(): SQL error: [MySQL][ODBC 3.51 Driver]Unknown database 'promodb',SQL state S1T00 in SQLConnect in fileMaintMaster.php The line of code that generates this error is: $access = odbc_connect('PromoDB','root','');
View Replies !
Warning: Undefined Index: 2 In
wat does this error means? Warning: Undefined index: 2 in C:Inetpubwwwroot emplateadmin.php on line 453 if i manually install php,(setting manually), this error wil not appear. but if i use the .exe to install, (means i dont' have to do the settings), this error will appear.
View Replies !
$http_post_vars (Warning: Undefined Index: 23232)
I have a form with one text box and when I submit the form and try and call the value on the next php page like this echo $HTTP_POST_VARS["myIP"]; I get the following error: Warning: Undefined index: 23232 in D:phpStatsdisplay_stats.php on line 6 however if I use the while loop to get the data from the form it works fine. Anyone know why this is happening?
View Replies !
Mysql_fetch_array() Error - Php Code Error
I have the following error on my php code: PHP Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result.This is part of the PHP where the problem is and the message i get on the HTML page is something wrong in artistinfo table: Code:
View Replies !
|