CODEWITHSUNDEEP
phpmysqlnull
bagofmilk
bagofmilk

Reputation: 1550

MYSQL PHP PDO Select Statement calling null

I'm trying to get the user's ID number from my table. For some reason the value always comes up "NULL", and it shouldn't be but I can't figure out what I'm doing wrong here.

Here is how my table 'users' looks:

enter image description here

<?php
  ...........................................
  ....connection details (connection is not the problem as other SQL queries in my code work fine)
  ...........................................
  ...........................................

  $getvals = $db->prepare("SELECT MFG_LINE, PM_MECHANICAL, PM_DESIGN, PM_APPLICATIONS, PM_PROGRAM, DESCRIPTION FROM new_schedule WHERE ITEM = '$jobnum'");
    $getvals->execute();
    while ($row = $getvals->fetch(PDO::FETCH_ASSOC)){
        $prod_line=$row["MFG_LINE"];
        $pe=$row["PM_MECHANICAL"];
        $de=$row["PM_DESIGN"];
        $ae=$row["PM_APPLICATIONS"];
        $ce=$row["PM_PROGRAM"];
        $model=$row["DESCRIPTION"];
    }



  /*Is job PE, DE, or CE?*/
    $engtype = rand(1,3);

    if ($engtype===1) { $engineer = $pe; }
    else if ($engtype===2) { $engineer = $de; }
    else if ($engtype===3) { $engineer = $ce; }
    else { $engineer = "Error 1005"; }

    echo $engineer;

    if ($engineer == null || $engineer = "") {
        $theengineer = 0;
        echo "nope";
    } else {        
        $getidnum = $db->prepare("SELECT USERID FROM users WHERE fullname LIKE '$engineer'");
        $getidnum->execute();
        $getthenum = $getidnum->fetch(PDO::FETCH_ASSOC);
        $theengineer = $getthenum['USERID'];
    }      

?>

The value is returning a NULL value when it should be returning "12". What am I missing here?

Upvotes: 0

Views: 46

Answers (3)

Dave Walker
Dave Walker

Reputation: 122

It maybe that there is spaces where not expected, so your query fails.

You need to add a wildcard front and end and the middle of the fullname. So replace the space, or spaces (could be several, never trust user input), in the middle with "%". So you have this:

$engineer = "%BRAD%DAVIS%";

Try that and let me know if it works?

Upvotes: 0

user1720897
user1720897

Reputation: 1246

In your code, the line that is
$getidnum = $db->prepare("SELECT USERID FROM users WHERE fullname LIKE '$engineer'");.
Change that to
$getidnum = $db->prepare("SELECT USERID FROM users WHERE fullname LIKE $engineer");

Upvotes: 0

mister martin
mister martin

Reputation: 6252

Based on the comments, give this a try:

$getidnum = $db->prepare("SELECT USERID FROM users WHERE fullname LIKE :engineer");
$getidnum->execute(array(':engineer' => $engineer));

Upvotes: 1

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