i am not getting the value in $sql2. so can anyone please....
thank u
$sql2="select allocatedmemory from projects where projectname='$projectassign'";
mysql_error();
$data1=mysql_query($sql2);
echo '$data1';
i am not getting the value in $sql2. so can anyone please....
thank u
$sql2="select allocatedmemory from projects where projectname='$projectassign'";
mysql_error();
$data1=mysql_query($sql2);
echo '$data1';
Jump to PostTry this.
$sql2="select allocatedmemory from projects where projectname='$projectassign'"; $data1=mysql_query($sql2) or die(mysql_error()); echo '$data1';
[edit]
Oops, and the echo will need changing to suit your purposes. What are you trying to do on the echo line, trying to display allocatedmemory? If so then try the following
Jump to PostWarning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 6 in C:\Program Files\Apache Group\Apache2\htdocs\Project\upload.php on line 50
$data1Yes I received your pm and the reason for that bug would be because the query returns zero results (no rows). So to solve this use the following:
…
Try this.
$sql2="select allocatedmemory from projects where projectname='$projectassign'";
$data1=mysql_query($sql2) or die(mysql_error());
echo '$data1';
[edit]
Oops, and the echo will need changing to suit your purposes. What are you trying to do on the echo line, trying to display allocatedmemory? If so then try the following
$sql2="select allocatedmemory from projects where projectname='$projectassign'";
$result=mysql_query($sql2) or die(mysql_error());
$data1=mysql_result($result,0);
echo $data1;
[/edit]
Warning: mysql_result() [function.mysql-result]: Unable to jump to row 0 on MySQL result index 6 in C:\Program Files\Apache Group\Apache2\htdocs\Project\upload.php on line 50
$data1
Yes I received your pm and the reason for that bug would be because the query returns zero results (no rows). So to solve this use the following:
$sql2="select allocatedmemory from projects where projectname='$projectassign'";
$result=mysql_query($sql2) or die(mysql_error());
if (mysql_num_rows($result)>0) {
$data1=mysql_result($result,0);
} else {
$data1='No data.';
}
echo $data1;
hey if i executed the same sql in mysql i am getting the value. but now it is displaying No Data.
Perhaps $projectassign isn't it's expected value.
Perhaps $projectassign isn't it's expected value.
i think the error might be here
$data1=mysql_result($d,0);
i think the error might be here
$data1=mysql_result($d,0);
Did you read my previous example fully as that code has a bug in it if your following my example. Not that that particular line would effect the if statement.
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