how to show every data from a table and how to run a 2 query in while loop?

Question

<?php
        session_start();
        include("dbconnect.php");
        $sql="select * from $_POST[select_catalog_query]";
        $rslt=mysql_query($sql);


        if(!mysql_fetch_array($rslt))
        {
            echo "<script type='text/javascript'>\n";
            echo "alert('There is No Data');\n";
            echo "</script>";
            echo "<script>document.location.href='admin.php'</script>";

        }
        while($row = mysql_fetch_array($rslt))
        {
            echo "<tr>";
            echo "<td>$row[id]</td>";
            $sql1 = "select * from login WHERE id=$row[id]";
            $rs = mysql_query($sql1);
            $rw1 = mysql_fetch_array($rs);
            echo "<td>$row[firstName]</td>";
            echo "<td>$row[lastName]</td>";
            echo "<td>$row[gender]</td>";
            echo "<td>$row[phone]</td>";
            echo "<td>$row[email]</td>";
            echo "<td>$rw1[type]</td>";
            echo "<td>$rw1[status]</td>";
            echo "</tr>";
        }
?>

i have 3 data in my 1st table...bt in output it's showing only 2 data...the last inserted data is not showing. What is it's solution? And in while loop when i am executing 2nd query for login table, a warning is given like this mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\myfile\as1\selectAdmin.php on line 41
how do i execute 2nd query??

change `$rslt=mysql_query($sql);` to `$rslt=mysql_query($sql) or die(mysql_error());` and see what's the error if any — Vipin Kumar KM, Nov 06 '13 at 06:48

score 0 · Answer 1 · answered Nov 06 '13 at 06:46

0

You don't need to run multiple queries here. If you do, the time to run will increase with every new row you have in the table $_POST[select_catalog_query] by a constant factor, which is bad because this will become very inefficient for large datasets.

Instead use the following query:

SELECT * FROM tableName t1 LEFT JOIN login t2 WHERE t2.id = t1.id

This way you'll only ever need to execute 1 query.

answered Nov 06 '13 at 06:46

Matt Harrison

13,001
6
46
64

i know it's inefficient but i have to show all data from $_POST[select_catalog_query]..wht can i do? and there is also 1st problem remaining.. – saddam hossain Nov 06 '13 at 07:33
You can still use the query provided. – Matt Harrison Nov 06 '13 at 08:20

score 0 · Answer 2 · answered Nov 06 '13 at 06:56

0

$sql="
select
  c.*,
  l.*
FROM
  $_POST[select_catalog_query] as c
 LEFT JOIN
 login as l
WHERE
 l.id = c.id";

 $rslt=mysql_query($sql);       


 if(mysql_num_rows($rslt)==0)
 {
        echo "<script type='text/javascript'>\n";
        echo "alert('There is No Data');\n";
        echo "</script>";
        echo "<script>document.location.href='admin.php'</script>";

}
else
{
    while($row = mysql_fetch_array($rslt))
    {
       /*YOUR data*/
    }
}

answered Nov 06 '13 at 06:56

it's not working...There is No Data..showing – saddam hossain Nov 06 '13 at 08:13
what do you want to echo? – Nov 06 '13 at 08:27

score 0 · Answer 3 · answered Nov 07 '13 at 05:48

0

SELECT * FROM table INNER JOIN table2 ON table.ID=table2.ID;

in this when both id will become same then it will fetch all the data

answered Nov 07 '13 at 05:48

chirag

103
1
3
10

score -1 · Answer 4 · answered Nov 06 '13 at 06:42

-1

here you said required 1 as parameter it means mysql_fetch_array is getting null as parameter you should check database also

and your question is not clear ask properly

id=$row[id] should be id=$row['id'] try it now

answered Nov 06 '13 at 06:42

chirag

103
1
3
10

that will give a syntax error!! – saddam hossain Nov 06 '13 at 06:53

how to show every data from a table and how to run a 2 query in while loop?

4 Answers4