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Inserting values in mysql database

Posted on 2012-12-21

1 solution

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Last Modified: 2012-12-22

Hello Experts,

I would like to insert the values from the PHP page into MYSQL database where the values were in the form of ...

<?php 
 for ($i = 0; $i < count($_POST['chkbx']); $i++) {
            $chkd = $_POST['chkbx'][$i];
            $filename = $_POST['myfilename'][$i];
            $src = $_POST['mysource'][$i];
            $tgt = $_POST['mytarget'][$i];
            
}
?>

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For example, there are 5 rows on php and user has selected any of the rows...that row data from php needs to inserted into mysql database that many rows.

Can you suggest the query for this or any references?

Thanks,
Shail

Comment

Question by:ShaileshShinde

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7 Comments

LVL 17

Expert Comment

by:Kent Dyer

ID: 387151172012-12-21

You mean something like..

<?php
   $conn = mysql_connect ("localhost","login","password");
   $db = mysql_select_db('database',$conn);

$LIBRARY_NO=$_POST['LIBRARY_NO'];
$TITLE=mysql_real_escape_string($_POST['TITLE']);
$LYRICIST=$_POST['LYRICIST'];
$VOICING=$_POST['VOICING'];
$NO_OF_COPIES=$_POST['NO_OF_COPIES'];

// Update the database if form posted 
if ($_POST['cmdadd'] != null)
{ 
$sql = "INSERT INTO Music(";
if ($LIBRARY_NO != null) {$sql=$sql."`LIBRARY_NO`";}
if ($TITLE != null) {$sql=$sql.", `TITLE`";}
if ($LYRICIST != null) {$sql=$sql.", `LYRICIST`";}
if ($VOICING != null) {$sql=$sql.", `VOICING`";}
if ($NO_OF_COPIES != null) {$sql=$sql.", `NO_OF_COPIES`";}
$sql=$sql.")"; 
//echo $sql;
$result = mysql_query($sql) or die("SQL Update failed"); 
  if ( ! $result )
    die ("Couldn't update: ".mysql_error());
  print "<h1>Table updated ". mysql_affected_rows() . 
  " row(s) changed</h1><p>";
}
?>

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HTH,

Kent

LVL 31

Expert Comment

by:Marco Gasi

ID: 387151382012-12-22

<?php 
 for ($i = 0; $i < count($_POST['chkbx']); $i++) {
            $chkd = $_POST['chkbx'][$i];
            $filename = mysql_real_escape_string($_POST['myfilename'][$i]);
            $src = mysql_real_escape_string($_POST['mysource'][$i]);
            $tgt = mysql_real_escape_string($_POST['mytarget'][$i]);
            $query = "INSERT INTO table (filename, src, tgt) VALUES ('$filename', '$src', '$tgt')";
            //execution code here 
}
?>

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then execute the query with mysql, mysqli or PDO. Keep in mind that now the use of mysql is discouraged and it is recommended the use of mysqli or PDO

If you want, you can post the markup of the form so I can be a bit more precise.

Cheers

LVL 1

Author Comment

by:ShaileshShinde

ID: 387151492012-12-22

Hello Expert,

The markup is something like this...

echo "<tr class=\"d" . ($i & 1) . "\">
                        <td><input class=textboxa type=text style=\"font-style: Italic;font-weight: bold;\" value=\"$i\" /><br/></td>
                        <td><input name=test_$i class=textboxc type=text style=\"font-style: Italic;font-weight: bold;\" value=\"" . $row['Delivery'] . "\" title=Delivery /><br/></td>
                        <td><input name=subtest_$i class=textboxc type=text style=\"font-style: Italic;font-weight: bold;\" value=\"" . $row['quality'] . "\" title=quality /><br/></td></tr>";

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LVL 31

Expert Comment

by:Marco Gasi

ID: 387151822012-12-22

I don't see checkboxes: how user selects rows? In your first snipet you used $_POST['chkbx'] so I presumed you had a checkbox per row in your form...

LVL 1

Author Comment

by:ShaileshShinde

ID: 387152002012-12-22

Hello expert,

Yes, there is an checkbox each row...

<td><input name=chk[] type=checkbox value=$i /><br/></td>

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LVL 31

Accepted Solution

by:Marco Gasi

Marco Gasi earned 400 total points

ID: 387152212012-12-22

Well, then you have to use 'chk' instead of 'chkbx':

<?php 
 for ($i = 0; $i < count($_POST['chk']); $i++) {
            $chkd = $_POST['chkbx'][$i];
            $filename = mysql_real_escape_string($_POST['myfilename'][$i]);
            $src = mysql_real_escape_string($_POST['mysource'][$i]);
            $tgt = mysql_real_escape_string($_POST['mytarget'][$i]);
            $query = "INSERT INTO table (filename, src, tgt) VALUES ('$filename', '$src', '$tgt')";
            mysql_query($query) or die(mysql_error()); 
}
?>

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LVL 1

Author Closing Comment

by:ShaileshShinde

ID: 387154452012-12-22

Thanks Experts!

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