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Views: 469 | Replies: 16
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Join Date: Jul 2008
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Newbie Poster
I have very limited use of using MySql - but I am trying to fix this code and it seems to be corrent. But I am still getting the error.
The website is a web directory, and the code itself checks to see the websites that arent linking back to the site.
This is Line 34
This is Lines 32 - 35
I have searched the net and tried various other methods that were posted on this forum and others and still I get the same error.
Any help and expertise would be much appreciated.
The website is a web directory, and the code itself checks to see the websites that arent linking back to the site.
Warning: mysql_num_rows(): supplied argument is not a valid MySQL result resource in /home/imedia/public_html/uk-webdirectory.info/admin/linkchecker.php on line 34
This is Line 34
language Syntax (Toggle Plain Text)
This is Lines 32 - 35
language Syntax (Toggle Plain Text)
I have searched the net and tried various other methods that were posted on this forum and others and still I get the same error.
Any help and expertise would be much appreciated.
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Join Date: Nov 2007
Location: Bangalore, India
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The error is because there is something wrong in your query. Print the query, execute it in mysql console or phpmyadmin. (Also post it here) 
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
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*PM asking for help will be ignored*
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Join Date: Jul 2008
Location: Hyderabad,India.
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Posting Pro
change this :
$return=mysql_query($query,$link);
to:
$return=mysql_query($query);
$return=mysql_query($query,$link);
to:
$return=mysql_query($query);
Champions Are Not SuperNatural,They Just Fought For One More Second When Every One Else Quit...Some times That One Second Of Effort Gives You The VICTORY...
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Join Date: Nov 2007
Location: Bangalore, India
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Originally Posted by Shanti Chepuru 
change this :
$return=mysql_query($query,$link);
to:
$return=mysql_query($query);
That definitely wouldn't matter. Thats because mysql_query can take an optional parameter of link identifier (the variable used to 'open' the connection).
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
*PM asking for help will be ignored*
*PM asking for help will be ignored*
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Join Date: Jul 2008
Location: Hyderabad,India.
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Posting Pro
k...nav...
thanks for indiacating........
thanks for indiacating........
Champions Are Not SuperNatural,They Just Fought For One More Second When Every One Else Quit...Some times That One Second Of Effort Gives You The VICTORY...
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Join Date: Jul 2008
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Newbie Poster
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Originally Posted by nav33n
The error is because there is something wrong in your query. Print the query, execute it in mysql console or phpmyadmin. (Also post it here)
I am not sure what you mean by printing the query, but I did test the first line and then the group of lines and the results were.
Tested:
$total=mysql_num_rows($return);
MySQL said: #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$TOTAL=mysql_num_rows($return)' at line 1
Tested:
$query="SELECT * FROM dir_site_list WHERE site_sponsor='N' AND site_live='Y'ORDER BY site_id LIMIT $start,$limit";
$return=mysql_query($query,$link);
$total=mysql_num_rows($return);
$sess_id="PHPSESSID=".session_id();
MySQL said: Documentation #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query="SELECT * FROM dir_site_list WHERE site_sponsor='N' AND site_live='Y'ORDE' at line 1
Where can I find the manual and version of MySQL I have and any ideas on how to fix it. I really appreciate anything you can tell me, as I am a newbie at this. Usually I just install scripts, not edit them.
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Join Date: Nov 2007
Location: Bangalore, India
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php Syntax (Toggle Plain Text)
Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
*PM asking for help will be ignored*
*PM asking for help will be ignored*
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Join Date: Jul 2008
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Newbie Poster
It's reply is practically the same as before I think.
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query="SELECT * FROM dir_site_list WHERE site_sponsor='N' AND site_live='Y' ORD' at line 1
You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '$query="SELECT * FROM dir_site_list WHERE site_sponsor='N' AND site_live='Y' ORD' at line 1
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Join Date: Nov 2007
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Can you post your updated code ?
echo $query; should print out the query. But anyway, post your complete code. Programming today is a race between software engineers striving to build bigger and better idiot-proof programs, and the Universe trying to produce bigger and better idiots. So far, the Universe is winning.
*PM asking for help will be ignored*
*PM asking for help will be ignored*
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Join Date: Jul 2008
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Newbie Poster
What do you mean by updated code? Also I am confused about print out the query, do you mean its response to the query I run in phpmyadmin?
Did you want me to post the whole page?
Did you want me to post the whole page?
Last edited by missaaliyah : 28 Days Ago at 7:02 am.
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