vchris Posted November 7, 2007 Share Posted November 7, 2007 I'm trying to create a subquery but it doesn't seem to be working.$sql1 = "SELECT * FROM db1.table1 WHERE field1=(SELECT post_id FROM db1.table2 WHERE field1=9795)";$result1 = mysql_query($sql1,$dbc1);while($row1 = mysql_fetch_assoc($result1)){ echo $row1['field1'].'<br />';}I get this error:Warning: mysql_fetch_assoc(): supplied argument is not a valid MySQL result resource in...Both queries work fine by themselves and looks to me like the subquery syntax is fine. Anything I could be missing? Link to comment Share on other sites More sharing options...
Ingolme Posted November 7, 2007 Share Posted November 7, 2007 Maybe you should do the other query first and add it into the second query after. Like this:$res0 = mysql_query("SELECT post_id FROM db1.table2 WHERE field1=9795",$dbc1);$row0 = mysql_fetch_array($res0);$sql1 = "SELECT * FROM db1.table1 WHERE field1=(".$row0['post_id'].")";$result1 = mysql_query($sql1,$dbc1);while($row1 = mysql_fetch_assoc($result1)){echo $row1['field1'].'<br />';} Link to comment Share on other sites More sharing options...
vchris Posted November 7, 2007 Author Share Posted November 7, 2007 But why would I have to do that? That's why they invented subqueries. Link to comment Share on other sites More sharing options...
Ingolme Posted November 7, 2007 Share Posted November 7, 2007 I don't know, I'm just suggesting because I've never used subqueries, so I can't be sure exactly how they work. If performing the query before works ok then at least your script will be working.Maybe somebody else can find a different solution. Link to comment Share on other sites More sharing options...
Synook Posted November 7, 2007 Share Posted November 7, 2007 Trying echoing mysql_error() after performing the query. That will tell you what is wrong. Link to comment Share on other sites More sharing options...
justsomeguy Posted November 8, 2007 Share Posted November 8, 2007 You guys are funny, just trying stuff without looking up the syntax first. You don't use the equal sign with subqueries, you use IN.$sql1 = "SELECT * FROM db1.table1 WHERE field1 IN (SELECT post_id FROM db1.table2 WHERE field1=9795)";Also, make sure that the version of MySQL you're using supports subqueries. I believe they added that in version 4. If not, you can use a join instead.$sql1 = "SELECT db1.table1.* FROM db1.table1, db1.table2 WHERE db1.table1.field1=db1.table2.post_id AND db1.table2.field1=9795"; Link to comment Share on other sites More sharing options...
SpOrTsDuDe.Reese Posted November 8, 2007 Share Posted November 8, 2007 Yet AGAIN. Justsomeguy has the correct answer. Link to comment Share on other sites More sharing options...
vchris Posted November 9, 2007 Author Share Posted November 9, 2007 Works! Thank you. Link to comment Share on other sites More sharing options...
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