Can't Understand Why This Code is'nt Working

[JamesThePanda]

Hi

 

I cant understand why this code isnt echoing the results.

 

$ids = $_POST['uid'];

$con = mysql_connect("localhost","root","");

mysql_select_db("product", $con);

$sqlout = mysql_query("SELECT * FROM product2 WHERE 'ID' IN ('$ids')");

while ($sqlres = mysql_fetch_assoc($sqlout))
  {
  echo $sqlres['filename'] . " " . $sqlres['title'];
  }

mysql_close($con);

 

 

Anyone got any ideas?

 

Thanks

 

James

[AyKay47]

There is an error in your SQL, if you had proper error handling established you would know this.

The only way that double quotes can be used as an identifier quoting character is when the SQL mode include ANSI_QUOTES. Either way you are using single quotes to encase the field `id`.

Use either back-ticks or none at all.

Also, to use the IN clause here, $ids would need to be a comma delimited list of values, which I doubt it is.

 

$ids = $_POST['uid'];

$con = mysql_connect("localhost","root","");

mysql_select_db("product", $con);

$ids = intval($ids); //sanitize data
$sqlout = mysql_query("SELECT * FROM product2 WHERE ID = '$ids'") or die(mysql_error()); //select only the fields that you are going to use

while ($sqlres = mysql_fetch_assoc($sqlout))
{
    echo $sqlres['filename'] . " " . $sqlres['title'];
}

mysql_close($con);

[Psycho]

Easy - there were no results. One reason my be that you put the field name in single strait quotes. So instead of looking for any records were the field ID has a value in the list, the query is trying to see if the STRING 'ID' is in the list of values. You should create your queries as a string variable so you can echo to the page for debugging purposes. And you can add other code to verify if there were results or not.

 

Also, I don't know what you expect the post value to look like, but you may not get the results you expect based on how it is formatted

$ids = $_POST['uid'];

$con = mysql_connect("localhost","root","");
mysql_select_db("product", $con);

$query = "SELECT `filename`, `title` FROM `product2` WHERE `ID` IN ('$ids')"
$result = mysql_query($query);

if(!mysql_num_rows($result))
{
    echo "There were no results";
}
else
{
    while($row = mysql_fetch_assoc($result))
    {
        echo "{$row['filename']} {$row['title']}<br>\n";
    }
}
mysql_close($con);

[JamesThePanda]

hmmm now its only returning a single result

 

I have a group checklist HTML form that were the array is started. I wanted the script to pull details from the database when I select certain items. now if I sect multiple items it only returns one result. 

[AyKay47]

make sure the checkbox input names are in array format.

Post the relevant code please.

[JamesThePanda]

ok this is what I have but for some reason its stoped working geting a parse error on the last line 

 

 

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
if (!isset($_POST['id'])){
mysql_select_db("product", $con);

$result = mysql_query("SELECT * FROM  product2");

echo'<form method="post" action="pro.php">';
while($row = mysql_fetch_array($result))
  {
  echo "<tr><td><input type=\"checkbox\" name=\"uid[]\" value=\"".$row['ID']."\" /></td><td>".$row['filename'] . "</td><td> " . $row['title']. "</tr> ";
  }

echo '<input type=\"submit\" name=\"formSubmit\" value=\"Submit\" />';
echo '</form>';
mysql_close($con);
?>

[AyKay47]

double quotes don't need escaped when the string is wrapped in single quotes, the backslashes will be interpolated as literal backslashes.

 

echo '<input type="submit" name="formSubmit" value="Submit" />';

[JamesThePanda]

ok here is the new "improved" code.

 

Also I'm getting a parse error on line 16

 

 

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
if (!isset($_POST['id'])){
mysql_select_db("product", $con);

$result = mysql_query("SELECT * FROM  product2");

echo'<form method="post" action="pro.php">';
while($row = mysql_fetch_array($result))
  {
  echo '<tr><td><input type="checkbox" name="uid[]" value=".$row['ID']." /></td><td>".$row['filename']."</td><td>".$row['title']."</tr>';
  }

echo '<input type="submit" name="formSubmit" value="Submit" /></form>';
?>

[AyKay47]

echo '<tr><td><input type="checkbox" name="uid[]" value="'.$row['ID'].'" /></td><td>"'.$row['filename'].'"</td><td>"'.$row['title'].'"</tr>';

[JamesThePanda]

Hi thanks for that

 

I replace that li with your code,

 

Now im getting an error on line 20

 

it makes no sence

 

<?php

$con = mysql_connect("localhost","root","");
if (!$con)
  {
  die('Could not connect: ' . mysql_error());
  }
if (!isset($_POST['id'])){
mysql_select_db("product", $con);

$result = mysql_query("SELECT * FROM  product2");

echo'<form method="post" action="pro.php">';
while($row = mysql_fetch_array($result))
  {
echo '<tr><td><input type="checkbox" name="uid[]" value="'.$row['ID'].'" /></td><td>"'.$row['filename'].'"</td><td>"'.$row['title'].'"</tr>';
}

echo '<input type="submit" name="formSubmit" value="Submit" /></form>';
?>

 

Thanks

 

James

[AyKay47]

instead of saying "i'm getting an error", please post the error and the line(s) that it refers to.

[homer.favenir]

you forgot the closing bracket of

if (!isset($_POST['id'])){