Warning: mysql_fetch_array(): supplied argument not a valid

How to use phpmyadmin.... phpmyadmin related arguments. Even phpMyAdmin issues

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DoppyNL

Wed Nov 06, 2002 6:53 am

This is an english forum, please post your question in english then we may be able to answer your question.


what I can conlude from the (englisch) error message is that your query failed. check the syntax of the query you executed.

Greetz Daan

/me thinks "Funny how french people allways think that everyone speaks french."

Jay

Wed Nov 06, 2002 8:28 am

DoppyNL wrote:This is an english forum, please post your question in english then we may be able to answer your question.


what I can conlude from the (englisch) error message is that your query failed. check the syntax of the query you executed.

Greetz Daan

/me thinks "Funny how french people allways think that everyone speaks french."

/me think it's funny how people who can speak english can't spell it :D :wink:

Joel

Fri Mar 28, 2003 3:00 am

Yeah, I'm getting that same trouble. I think my error is in this part...

$first = $_GET["first"];

if ($first) {

$result = mysql_query("SELECT * FROM employees WHERE first=$first",$db);


$myrow = mysql_fetch_array($result);

can anyone see an error in that?

Renee

Tue Apr 15, 2003 7:22 am

check the value of $first. If it is alright, add single quote to $first in the query and try.

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bezmond
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Tue Apr 15, 2003 7:48 am

$result = mysql_query("SELECT * FROM employees WHERE first='$first'");

that should work, if not...
echo $first;

check that $first as been set ok...

Andrew

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Redcircle
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Thu Jul 31, 2003 11:17 pm

check to make sure your mysql statement is valid and there are no misspelled words.

sigix
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Mon Aug 04, 2003 4:18 am

try to echo all the values (major on which your query is working) on the next page then you can easily spot the problem
if you are not getting the values then there is problem in your receiving end ...
get values with single quot $name=$HTTP_POSTVARS[''];;;; :arrow:

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