Solved

Inserting values in mysql database

Posted on 2012-12-21
7
465 Views
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
0
Comment
Question by:ShaileshShinde
  • 3
  • 3
7 Comments
 
LVL 17

Expert Comment

by:Kent Dyer
ID: 38715117
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
0
 
LVL 31

Expert Comment

by:Marco Gasi
ID: 38715138
<?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
0
 
LVL 1

Author Comment

by:ShaileshShinde
ID: 38715149
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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0
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LVL 31

Expert Comment

by:Marco Gasi
ID: 38715182
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...
0
 
LVL 1

Author Comment

by:ShaileshShinde
ID: 38715200
Hello expert,

Yes, there is an checkbox each row...

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

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0
 
LVL 31

Accepted Solution

by:
Marco Gasi earned 100 total points
ID: 38715221
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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0
 
LVL 1

Author Closing Comment

by:ShaileshShinde
ID: 38715445
Thanks Experts!
0

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