MYSQL PHP PDO Select Statement calling null

Question

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?

If you run SELECT USERID FROM users WHERE fullname LIKE 'BRAD DAVIS' from a MySQL client, whats the output? — user1720897
– user1720897, Commented May 1, 2014 at 19:17
hmm...when I do it that way it works. So I guess I need to update how I actually get my $engineer value — bagofmilk
– bagofmilk, Commented May 1, 2014 at 19:21
Yes indeed. You should post the code that you actually used, not code that is mostly the same as your actual code, but modified a little bit before you post it. That's like having a problem with your car which is making a funny noise, and taking it to the mechanic, but instead of bring the problem car, you bring one similar to the problem car. — Andy Lester
– Andy Lester, Commented May 1, 2014 at 19:25

mister martin · Accepted Answer · 2014-05-01 19:23:35Z

1

Based on the comments, give this a try:

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

answered May 1, 2014 at 19:23

mister martin

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6 Comments

bagofmilk Over a year ago

how do i retrieve the value from the array?

mister martin Over a year ago

You define the value before the query, just as you're currently doing.

mister martin Over a year ago

Then what are you not showing us? If the variable $engineer contains only "BRAD DAVIS" and nothing else, then there is nothing wrong with the query.

bagofmilk Over a year ago

I added more code (see above). I'm pulling engineer names from another jobs table. Then choosing a random position ($de,$pe,$ce). The engineer will be the random chosen one. All names are in the user database. I echo the engineer's name to make sure it is pulling a name and it always does.

bagofmilk Over a year ago

Ok...its stupid...but if you look at the "if" statement above the SQL call, I only had one "=" in there which was tripping it up yet not giving me the error.

|

user1720897 · Accepted Answer · 2014-05-01 19:26:27Z

0

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");

answered May 1, 2014 at 19:26

user1720897

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2 Comments

bagofmilk Over a year ago

I still get a null value.

user1720897 Over a year ago

Could you also echo the resulting SQL query.

Dave Walker · Accepted Answer · 2014-05-01 19:27:39Z

0

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?

answered May 1, 2014 at 19:27

Dave Walker

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