Query Result = Resource ID #10
I have an array $countertop.... i run through each one like this.
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Another Invalid Result Resource
Sorry to ask a debugging question, but I've been fooling with this thing all day, and I can't see what the problem is. $query is passed from search form: $result = mysql_query("select id, date, title, text, reviewer, email, score, cover, url, url_title, hits, type, creator, release, upc, quote from one7_reviews where title LIKE '%$query%' OR description LIKE '%$query%' OR creator LIKE '%$query%' OR url_title LIKE '%$query%' ORDER BY title ASC"); $num = mysql_num_rows($result); I keep getting "Supplied argument is not a valid MySQL result resource "
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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.
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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"
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Fetch_array: Invalid Result Resource
Any suggestion as to why my fetch_array is bad? Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/jkokko/public_html/ask/test.php on line 11 ******************* <?php $dbHost = "localhost"; $dbUser = "user"; $dbPass = "pass"; $dbDatabase = "dbname"; //connet to the database $db = mysql_connect("$dbHost", "$dbUser", "$dbPass") or die ("Error connecting to database."); mysql_select_db("$dbDatabase", $db) or die ("Couldn't select the database."); $result = "SELECT a_uid FROM `answers` WHERE `qid` = ཉ' LIMIT 0, 30, $db"; while ($r = mysql_fetch_array($result)) { echo $r; } ?> ****************
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Not A Valid MySQL Result Resource
I have a website which i have a login script for however I want to users to be able to have an account balance. I have set up a field for balance in the users table of my database however I am having problems fetching the correct row for each user. Code:
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Mysql_num_rows(): Not A Valid MySQL Result Resource?
i'm getting three unusual errors which i can't figure out how to correct-- or even necessarily what's causing them. (note: not that you'd 'count' the lines, but these numbers might not match the code below as i've done some editing in between when i started this post and now) Code:
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Supplied Argument Is Not A Valid MySQL Result Resource
Before I get in trouble, I have searched extensively for this one, in the PHP Docs, online etc. I have a simple page: <?php $un="jim"; $pw="jim"; $db="localhost"; $link = mysql_connect($db,$un,$pw) or die("Could not connect: " . mysql_error()); $sql2 = "SELECT * FROM eib.book ORDER BY bookid"; $op = mysql_query($sql2) ; $num= mysql_numrows($op); ?> When I run the page I keep getting this error: Warning: mysql_numrows(): supplied argument is not a valid MySQL result resource in C:Program FilesApache GroupApache2htdocseibphpitemadd2.php on line 34
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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 = $username; session_register("valid_user"); } } ?>
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Mysql_fetch_array :: Supplied Argument Is Not A Valid MySQL Result Resource
I've got a problem with mysql_fetch_array function. I have two function: function db_query($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".
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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 $db_usemap;} ?>
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Mysql_free_result Return Error Saying Not Valid Mysql Result Resource
im trying to do a query which takes rows of large amounts of text data and displays them one after another. the size of the fields is causing the page to be massive in size. so to try and reduce this im trying to use mysql_free_result to clear the contents from memory. however its returning an error saying not valid mysql data when im certain that it is. is this because im using a while loop like this...? ($row = mysql_fetch_result($result)) { mysql_free_result($result); }
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Mysql_num_rows(): Supplied Argument Is Not A Valid MySQL Result Resource
I've got the following error: Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/wbdfdart/public_html/wbdfforum/verwijder.php on line 13 verwijder.php 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 <?php include "connect.php"; if (isset ($cat) AND isset ($topic) AND isset ($message)) { $query = "SELECT user FROM reply WHERE id='$message'"; $exec = mysql_query($query); $result = mysql_fetch_array($exec); $owner = $result["user"]; $moderator_check_query = "SELECT * FROM moderator WHERE category=$cat AND user=$myid"; $moderator_check_exec = mysql_query($moderator_check_query); $moderator_result = mysql_num_rows($moderator_check_exec); if ($moderator_result == 1 OR $owner == $myid) { $moderator = 1; } else { $moderator = 0; } if ($moderator == 1) { $replydelete_query="DELETE FROM reply WHERE id ='$id'"; mysql_query($replydelete_query) or die(mysql_error()); $topicmineen_query="UPDATE topic SET replies='replies -1' WHERE id='$topic'"; mysql_query($topicmineen_query) or die(mysql_error()); echo "<script>location.href='index.php?cat=".$cat."&topic=".$topic."&page=".$page."'</script>"; } else { echo "<script>location.href='index.php?cat=".$cat."&topic=".$topic."&page=".$page."'</script>"; }} ?>
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Ldap_get_values() Stating That The Supplied Resource Is Not A Valid Ldap Result
I'm checking a user identity on a secure LDAP server using the following code: $ldapconn = ldap_connect("ldaps://myserver.mycompany.ch", 636 ) or die( "Can't connect to LDAP" ) ; $ldapresult = ldap_search( $ldapconn,"o=mycompany,c=ch","cn=".$name); if( $ldapresult ) { $entries = ldap_get_entries( $ldapconn, $ldapresult ) ; if( $entries["count"] ) { $ldapbind = ldap_bind( $ldapconn, $entries[0]['dn'], $pwd ) ; if( $ldapbind ) { echo "Connected: user exists, password checked<br>" ; } } This works ok if I submit the right $name and $pwd Then I'd like to retrieve some data from the specified user, say the givenName and sn but can't figure out a way to retrieve those values using ldap_read or ldap_get_values For example: $results = ldap_read($ldapconn, "cn=myname,o=mycompany,c=ch", '(objectclass=person)', array( "givenname", "sn" ) ) ; $firstname = ldap_get_values($ldapconn, $results, "givenname" ) ; $lastname = ldap_get_values($ldapconn, $results, "sn" ) ; This fails on the ldap_get_values() stating that the supplied resource is not a valid ldap result entry resource. If I use something like: $firstname = ldap_get_values($ldapconn, $ldapresult, "givenname" ) ; $lastname = ldap_get_values($ldapconn, $ldapresult, "sn" ) ; It fails too...
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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); ............
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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 "); } ?>
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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)
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Turn Select Query Result Into Hyperlink To Other Query
I have a query which gives results of selecting coursenames from a table called trainingtopics but this does so in a continuous bulk of text without any breaks between each record. my question how do I create line breaks between each record and also how can I force each query result to become a hyperlink which when clicked runs another query that gives details of that course. <?php $user = "root"; $host = "localhost"; $password = ""; $connection = mysql_connect($host, $root, $password) or die ("Couldn't connect to server."); $database = "courses"; $db = mysql_select_db($database) or die ("Couldn't select database."); $sql = "SELECT coursename FROM trainingtopics"; echo $sql."=sql<br>"; $result = mysql_query($sql) or die(mysql_error()); echo $result."=result<br>"; while($row=mysql_fetch_array($result)) { // NOTE this one ABOVE the echo echo "result found!"; echo $row[0]; } ?>
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MySQL Query Resource Usage
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"Supplied Argument Is Not A Valid MySQL Result Resource"
this is one of the first scripts im writing completely by myself but its driving me crazy. the errors its returning are: Quote: Warning: Supplied argument is not a valid MySQL result resource in c:apachehtdocsquiz_chap12_view.php on line 16 There are currently in the table: PHP Code:
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Deleting Records - "not A Valid Result Resource" Error.
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Use Query Result Field As Query Key For New Query..
$query  = "SELECT ProjID, UserID, ProjDesc, file, OrigProj, OrigUser, ProjDate FROM projects"; $result = mysql_query($query); I would like to take the field 'UserID', and utilise it for a Query statement to my users table in my database, to read and fill in information to my table, which looks like this: PHP Code:
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I dont know why this query is not returning a value, when I should be getting the value of ཋ'. PHP Code: mysql_select_db($dbname, $conn) or die (mysql_error()); $query = "SELECT MIN(q_id) FROM questions2 WHERE quiz_id='".$quiz_id."'"; echo $query.'<br />' $result = mysql_query($query) or die(mysql_error()); $row = mysql_fetch_array($result) or die(mysql_error()); $first = $row['q_id']; echo 'first='.$first.'<br />'
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Trying to have an echo display a result from a query. Not such a hard concept. Yet I am having trouble. What good is the data without analysis. Here is what I am using to try and grab result. Have been told it looks correct, must be missing something small. When this runs, alternate echo displays in browser. So you know: 'testdata' is the table 'score' & 'id' are columns or fields. Almost think it is missing a table structure to display in. Looking for player name column and score column (for the sake of example). Code:
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Displaying Result Of A Sum Query:
$sql = "SELECT sum(value) FROM `offers' WHERE sid='$sid'"; $result = mysql_query($sql); As you can see, I'm trying to select a column in my database which is named "value" I want to return the sum of the records which match the current user's subscriber id ($sid) I think the query is structured correctly. However, I'm having trouble placing the result of the query into a variable that php can echo to the page. In the database, the numbers are formatted as decimal. I just want to get the sum of the fields into a numeric php variable that I can add/subtract/divide etc and echo to the page. How do I go about doing this?
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