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Views: 8818 | Replies: 5
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Join Date: Oct 2005
Posts: 13
Reputation: 
Rep Power: 3
Solved Threads: 0
Newbie Poster
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
#1
Oct 16th, 2005
Hello,
I created this site on my local machine using PHP 5.0.4 and mysql 4.1 and it works perfectly fine.
I then ftp it to my webhost but the code to retrieve values give me this error.
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/laterallinks/ApplicantInformation.php on line 25
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/www/laterallinks/ApplicantInformation.php on line 621
I let my webhost know they so they upgrade from php 4 to php 5.0.5 but still use mysql client version 3.x
Here is the portion of the code that uses the functions.
segment 1:
$sql = mysql_query($query);
while($row = mysql_fetch_array($sql)){
?>
<div align="center">
<p align="right"><font face="Courier New"><font color="#0000FF"><b>Time of Application: </b></font> <? echo $row['dateApplied']?></font></p>
<table border="1" width= "750" style="border-collapse: collapse" id="table1" bgcolor="#F2FDFF">
segment 2:
$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM applicants ". $genQuery),0);
what could be possible wrong.
I created this site on my local machine using PHP 5.0.4 and mysql 4.1 and it works perfectly fine.
I then ftp it to my webhost but the code to retrieve values give me this error.
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource in /home/www/laterallinks/ApplicantInformation.php on line 25
Warning: mysql_result(): supplied argument is not a valid MySQL result resource in /home/www/laterallinks/ApplicantInformation.php on line 621
I let my webhost know they so they upgrade from php 4 to php 5.0.5 but still use mysql client version 3.x
Here is the portion of the code that uses the functions.
segment 1:
$sql = mysql_query($query);
while($row = mysql_fetch_array($sql)){
?>
<div align="center">
<p align="right"><font face="Courier New"><font color="#0000FF"><b>Time of Application: </b></font> <? echo $row['dateApplied']?></font></p>
<table border="1" width= "750" style="border-collapse: collapse" id="table1" bgcolor="#F2FDFF">
segment 2:
$total_results = mysql_result(mysql_query("SELECT COUNT(*) as Num FROM applicants ". $genQuery),0);
what could be possible wrong.
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Join Date: Apr 2005
Location: Auckland, New Zealand
Posts: 136
Reputation:
Rep Power: 4
Solved Threads: 1
Junior Poster
Re: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resou
#2
Oct 17th, 2005
It's probably not a php version problem
usually $sql and $query are the same thing and you use $result for the result set. That improves readability but doesn't fix anything
this should show where the problem is.
If it's the query paste it into phpMyAdmin and see the errors there. It may be something simple like a missed quote mark or a column in the table.
Sarah
usually $sql and $query are the same thing and you use $result for the result set. That improves readability but doesn't fix anything
$result = mysql_query($query) or die(mysql_error() . '<br />'. $query);
while($row = mysql_fetch_array($result)){
var_dump($row);
}this should show where the problem is.
If it's the query paste it into phpMyAdmin and see the errors there. It may be something simple like a missed quote mark or a column in the table.
Sarah
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Join Date: Oct 2005
Posts: 13
Reputation:
Rep Power: 3
Solved Threads: 0
Newbie Poster
Re: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
#3
Oct 17th, 2005
Sarah,
Thank so very much. but how can i be having problems online when the code works perfectly on my test machine.
Thank so very much. but how can i be having problems online when the code works perfectly on my test machine.
Last edited by iketunde : Oct 17th, 2005 at 5:22 am. Reason: typo error
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Join Date: Apr 2005
Location: Auckland, New Zealand
Posts: 136
Reputation:
Rep Power: 4
Solved Threads: 1
Junior Poster
Re: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
#4
Oct 17th, 2005
That points to a connection issue - but also to a potential difference in the structure of the tables. I was hoping the error message would give you one of those moments when you go "d'oh". It's frustrating but a solution is always a good thing.
We really need to know what mysql_error() returns.
We really need to know what mysql_error() returns.
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Join Date: Oct 2005
Posts: 13
Reputation:
Rep Power: 3
Solved Threads: 0
Newbie Poster
Re: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
#5
Oct 17th, 2005
Sarah...
thank u.. it worked. the its a case sensitive issue. i named the table Applicants and it refered to it as applicants;
thanks once more
thank u.. it worked. the its a case sensitive issue. i named the table Applicants and it refered to it as applicants;
thanks once more
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Join Date: Mar 2008
Posts: 1
Reputation:
Rep Power: 0
Solved Threads: 0
Newbie Poster
Re: Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
#6
Mar 17th, 2008
Warning: mysql_fetch_array(): supplied argument is not a valid MySQL result resource
code is here>>>>
$query = "SELECT * FROM upload_question_1 ORDER BY Rand() WHERE
exprience='".$expr."' AND gread='".$catg."' LIMIT 1";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
//echo "<tr><td>$row[0]</td><td>$row[1]</td><td>$row[2]
</td></tr>";
echo $row [1];
code is here>>>>
$query = "SELECT * FROM upload_question_1 ORDER BY Rand() WHERE
exprience='".$expr."' AND gread='".$catg."' LIMIT 1";
$result = mysql_query($query);
while ($row = mysql_fetch_array($result, MYSQL_NUM)) {
//echo "<tr><td>$row[0]</td><td>$row[1]</td><td>$row[2]
</td></tr>";
echo $row [1];
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