query select as input to next query

[devikacs]

I wanted to use the select of a mysql query as the input to the next mysql query.
The output of the first query is stored in $result.
Can I do select * from $result.
If not is there any other way.

[Lumpy]

I wanted to use the select of a mysql query as the input to the next mysql query.
The output of the first query is stored in $result.
Can I do select * from $result.
If not is there any other way.

To tell you for sure, I would need to see the code your working with for the first query, but it should work. Most likely you would set up your second query something like...

$query2 = "SELECT * from " . $result['fieldname'] . " any further query instructions here";
 
 

[adamalton]

Yeah that should be possible...[PHP]$query1="SELECT column_name FROM my_table WHERE column_name='sa usages'";
$result=mysql_q uery($query1);

$search_for = mysql_result($r esult,0,'column _name');//take the first result from query1

$query2 = "SELECT * FROM other_table WHERE other_column_na me='". $search_for ."'";
$result2 = mysql_query($qu ery2);

$final_result = mysql_result($r esult2,0,'other _column_name');[/PHP]

[Lumpy]

From looking at your code, I think you got it. I read through it a couple of times and I think it should work for you.

[devikacs]

<code=php>

$query="select * from search";
$result=mysql_q uery($query);
$num=mysql_numr ows($result);
echo $num;
$i=0;
while ($i < $num)
{
$name=mysql_res ult($result,$i, "name");
$rollno=mysql_r esult($result,$ i,"rollno");
echo"<p><b>$rol lno $name</b></p>";
$i++;
}

$year=3;
$query2="select * from ".$result."wher e year = ". $year;

$res=mysql_quer y($query2);
$num2=mysql_num rows($res);

echo $num2;
echo "something" ;
$i=0;
while ($i < $num2)
{
$name=mysql_res ult($res,$i,"na me");
$rollno=mysql_r esult($res,$i," rollno");
echo"<p><b>$rol lno $name</b></p>";
$i++;
}
</code>

This is what i have done. The first query gives proper output. But the second one doesnt print anything. The $num2 is also null. I do have values with year =3.
The php script is not generating any error.
What could possibly be wrong?