Member Avatar for yara1

hi

i design websit to view images


i want when admin click on the image image will be deleted


how i do that


??

this my code

<head>
<script type="text/javascript" >
function vvvv($id,$im)
{

if(confirm("are you sure want delet this imagej?"));

$b=true

if($b)
{
<?php


mysql_connect ('localhost', 'root');
mysql_select_db("pickapic");
$myquery="Delete FROM photo WHERE Pid=$id";
mysql_query($myquery);

?>

}


}

</script>
</head>

<body>







<?php

print("<table width=40% border=0 align=center>");

 mysql_connect ('localhost', 'root');
mysql_select_db("pickapic");

$result2= mysql_query("SELECT Pname FROM photo");
$result = mysql_query("SELECT Pid , link FROM photo");
$count=0;

while($row = mysql_fetch_array($result))
  {
  if ($count==0)
 print("<tr>");
 $image=$row["link"];
 $id=$row["link"];
 
 print ("<td><img src=\"file:///C|/xampp/htdocs/PickaPic/admin/img/$image \" width=\"150\" height=\"150\" border=\"0\" onClick=\"vvvv('$id','$image')\"; ></td>");

 $count=$count+1;
 
 if($count == 5 )
 {
 print("</tr>");
 print("<tr>");
 
for($i=0; $i<5 ; $i=$i+1)
{
$row2= mysql_fetch_array($result2);
$name1=$row2["Pname"];
print("<td><div align=\"center\"><span class=\"style6\"> $name1  </span></div></td>");


}
print("</tr>");
 $count=0;
 }
 

  }
 
  print("</table>");
  
 
?>


</body>

</html>

thx :):):)

  • 2 Contributors
  • 17 Replies
  • 86 Views
  • 1 Day Discussion Span
  • Latest Post Latest Post by yara1
Member Avatar for somedude3488

its easier to help when you explain what you want in more detail.

Member Avatar for yara1

hi kkeith

i explain it


and i have code i mix php with java script and it work as will

if that mistake

how i correct it please

Member Avatar for yara1

:(
:(
:(

Member Avatar for somedude3488

what are the smilies for?

Member Avatar for yara1

to tell i still have problem


plzzz i try many times

i dont sleep for 2 days

today is last day that means i must finish it today

i am really sad

Member Avatar for somedude3488

is it outputting the images ok, is the delete function the only thing you are having trouble with.

Member Avatar for somedude3488

i typed this up for you. i only fixed the delete images part and copied the rest of your code.

try it, and tell me if there are errors.

<?php

//Set page name
$thispage = 'index.php';

//Fill in database info
$host = 'localhost';
$user = 'root';
$pass = '';
$db   = 'pickapic';

$con = mysql_connect($host,$user,$pass) or die('Could not connect');
mysql_select_db($db) or die('Error: Could not select database');

if (isset($_GET['do'])) {
	$do = $_GET['do'];
	switch($do) {
		case "delete":
			if (isset($_GET['id'])) {
				$id    = mysql_real_escape_string($_GET['id']);
				$sql   = "DELETE FROM `photo` WHERE `Pid` = " . $id;
				$query = mysql_query($sql) or die('Error: ' . mysql_error());
			}
		break;
	}
}

$table  = '<table width="40%" border="0" align="center">';

$sql1   = "SELECT `Pname` FROM `photo`";
$sql2   = "SELECT `Pid`,`link` FROM `photo`";
$query1 = mysql_query($sql1);
$query2 = mysql_query($sql2);
$count  = 0;
while ($row = mysql_fetch_assoc($query1)) {
	if ($count == 0) {
		$table .= '<tr>';
	}
	$image = $row['link'];
	$id    = $row['link'];
	$table .= '<td><img src="file:///C|/xampp/htdocs/PickaPic/admin/img/' . $image . '" width="150" height="150" border="0" onClick="vvvv(\'' . $id . '\');" /></td>';
	$count++;
	if ($count == 5) {
		$table .= '</tr>';
		$table .= '<tr>';
		for ($i = 0; $i < 5; $i++) {
			$row2 = mysql_fetch_assoc($query2);
			$name1 = $row2['Pname'];
			$table .= '<td><div align="center"><span class="style6"> ' . $name1 . ' </span></div></td>';
		}
		$table .= '</tr>';
		$count  = 0;
	}
}
$table .= '</table>';

mysql_close($con);

?>
<html>
<head>
<title></title>
<script type="text/javascript">
function vvvv(id) {
	if (confirm("Are you sure you want to delete this image?")) {
		window.location = '<?php echo $thispage;?>?do=delete&id='+id;
	}
}
</script>
</head>
<body>
<?php echo $table;?>
</body>
</html>
Member Avatar for somedude3488

is this is personal project?

i was just curious

Member Avatar for yara1

hi

in my previouse code the alert appare

but code to delet image from database dosn excute

i mean image didnt deleted

no , i have another in my add photo code


alert dont appare i dont know why


??

can u fix it plzz?

Member Avatar for somedude3488

that was really confusing. are you saying that my confirm javascript didn't execute and the image was not deleted.

or are you talking about another code.

Member Avatar for yara1

:)


i said in my above code

alert appare but data dosnt remove from my database

in my delet code i think the error when i pass variable as prameter and recive it

becoz when i give $id value inside my function data removed

but when i recive $id as pramater
data dosnt remove

i have another code for adding

alert dosnot appare

but data add to my data base


but now i care much about my delet code

Member Avatar for somedude3488

i have no idea what the problem is. i would have to see it for myself to maybe find the error. i didn't see anything wrong with my code.

Member Avatar for yara1

can i send my code to u

Member Avatar for yara1

and sorry coz i late

Member Avatar for somedude3488

i guess you can send it to me. i won't be able to get to it until later today.

Member Avatar for yara1

Error: You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near '' at line 1


this error appare

Member Avatar for yara1

is this is personal project?

i was just curious

yes its

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