display user input value

Question

+-----+-------+--------+
  ID  | Owner | Number |
+-----+-------+--------+
 com1 | peter | 103045 |
+-----+-------+--------+
 com2 | deter | 103864 |
+-----+-------+--------+

Hey guys. I am working on this part of the project where the user is suppose to input a certain ID. When he input this certain ID it is suppose to print out everything in the row w that ID. In this case if he inputs com1, it will display:

+-----+-------+--------+
  ID  | Owner | Number |
+-----+-------+--------+
 com1 | peter | 103045 |
+-----+-------+--------+

I used this form and code:

form.php

    <form action="query.php" method="POST">
    Enter your ID: <input type="varchar" name="id" /><br />
    <input type="submit" value="Audit" />
    </form>

query.php

   $con = mysql_connect("localhost", "root", "");
    if (!$con)
        {
        die('Could not connect: ' . mysql_error());
        }

    mysql_select_db("livemigrationauditingdb", $con);


    $input = (int) $_POST['id'];
    $query = mysql_query("SELECT * FROM system_audit WHERE ID = $input");

    echo "<table border='1'>
    <tr>
    <th>ID</th>
    <th>Owner</th>
    <th>Number</th>
    </tr>";


    while($row = mysql_fetch_array($query))
    {
      echo "<tr>";
      echo "<td>" . $row['ID'] . "</td>";
      echo "<td>" . $row['Owner'] . "</td>";
      echo "<td>" . $row['Number'] . "</td>";
      echo "</tr>";
      }
    echo "</table>";

    mysql_close($con);

think you guys know the mistake? oh and is there away for me not to link it to other page? i need the form and codes to be in one page. currently my form links to query.php. is there a way to cancel this line and it still works. thanks!

use $input = $_POST['id'] as Rohan Sood suggests, when casting to an integer and there are characters in the string, the characters will be left out, therefor no ID will match com1 because it will result in just 0 when casting to an integer. — dbf
– dbf, Commented Sep 8, 2012 at 18:09
Put $input in single quotations like: "SELECT * FROM system_audit WHERE ID = '$input' " — Siamak Motlagh
– Siamak Motlagh, Commented Sep 8, 2012 at 18:11
hi siamak and dbf. thank you but i have tried both before. it returns the following message and shows no result Warning:mysql_fetch_array() expects parameter 1 to be resource, boolean given in C:\xampp\htdocs\query.php — user1629618
– user1629618, Commented Sep 8, 2012 at 18:12
@user1629618 this error happens when it's something wrong with the query string or mysql connection or database selection. make sure that you connected to mysql and select table before the query string and then check the table name "system_audi" and the col name "ID". — Siamak Motlagh
– Siamak Motlagh, Commented Sep 8, 2012 at 18:19
@user1629618 - try echoing out your generated query and entering it directly in the database. — andrewsi
– andrewsi, Commented Sep 8, 2012 at 18:19

Rohan Sood · Accepted Answer · 2012-09-08 18:07:02Z

1

Try removing the (int) from $input = (int) $_POST['id'];

answered Sep 8, 2012 at 18:07

Rohan Sood

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6 Comments

Rohan Sood Over a year ago

Please tell us the result for a print_r($row);

user1629618 Over a year ago

echo "<tr>"; echo "<td>" . $row['ID'] . "</td>"; echo "<td>" . $row['Type'] . "</td>"; echo "<td>" . $row['Setting'] . "</td>"; echo "<td>" . $row['Value'] . "</td>"; echo "<td>" . $row['Actual'] . "</td>"; echo "</tr>"; is just to put them in table form

Rohan Sood Over a year ago

I got that bro. Just in your document, query.php, put in a "print_r($row)" right after the opening braces for the while statement. And tell us the output.

user1629618 Over a year ago

OH! pardon me. i havent slept properly for a few days. and nah. it didnt work. says there's a systax error

Rohan Sood Over a year ago

What is your result for <?php echo "SELECT * FROM system_audit WHERE ID = $input";?>

|

Rahul Kadukar · Accepted Answer · 2012-09-08 18:20:33Z

0

Try removing

while($row = mysql_fetch_array($query))

and replace it by

while($row = mysql_fetch_row($result))

Also remove the (int) and see if it helps.

  echo "<td>" . $row['ID'] . "</td>";
  echo "<td>" . $row['Owner'] . "</td>";
  echo "<td>" . $row['Number'] . "</td>";

Can be replaced by

  echo "<td>" . $row[0] . "</td>";
  echo "<td>" . $row[1] . "</td>";
  echo "<td>" . $row[2] . "</td>";

You can also have both of them on the same page as shown

<?php

    $con = mysql_connect("localhost", "root", "");
    if (!$con)
        {
        die('Could not connect: ' . mysql_error());
        }

    mysql_select_db("livemigrationauditingdb", $con);


    $input = (int) $_POST['id'];

    if($input)
    {
      $query = mysql_query("SELECT * FROM system_audit WHERE ID = $input");

      echo "<table border='1'>
      <tr>
      <th>ID</th>
      <th>Owner</th>
      <th>Number</th>
      </tr>";


      while($row = mysql_fetch_row($query))
      {
        echo "<tr>";
        echo "<td>" . $row[0] . "</td>";
        echo "<td>" . $row[1] . "</td>";
        echo "<td>" . $row[2] . "</td>";
        echo "</tr>";
      }
      echo "</table>";
    }

    mysql_close($con); 
?>   

    <form action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST">
    Enter your ID: <input type="varchar" name="id" /><br />
    <input type="submit" value="Audit" />
    </form>

?>

edited Sep 8, 2012 at 18:20

answered Sep 8, 2012 at 18:12

Rahul Kadukar

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2 Comments

Rohan Sood Over a year ago

mysql_fetch_row is generally considered as a relatively better practice due to the increased readability of the code. So I wouldn't really recommend your approach, even though it might just get the work done.

user1629618 Over a year ago

hey again rohan. mm. you said to replace $query to $result. mm, the thing is i did not declare any $result

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