PHP MySQL Where clause isn't working

Question

I have a database like which has multiple columns and when querying it with a WHERE clause it won't get any results. Here is the code I am using :

$columns = $_GET['var'];
$where = $_GET['where'];
$checkValue = $_GET['checkValue'];
$userInput = $_GET['userInput'];

$query = "SELECT ";

foreach($columns as $val)
    $query .= "$val, ";

$query .= "FROM Email";

if($where === "yes")
    $query .= " WHERE $checkValue = '$userInput'";

$columns is multiple checkboxes for the user to select which columns they wish to see. It works perfectly except when adding the where clause. When I've been testing it I made sure that the it was exactly the same as in the database. Also the $checkValue is a dropdown list which values are exactly the same as in the database. Also just to note later on I edit the query so the last comma is removed.

To print it out I use :

while($c = mysqli_fetch_assoc($results)){
    foreach($columns as $val){
        $header = ucwords($val);
        echo "<b>$header</b><br>";
        echo $c[$val]."<br>";   
    }
    echo "-------------------------------<br>";
}

This is the query that is outputted when not using the where clause and works:

 SELECT date, mediatype FROM Email

And here is the query that doesnt work:

 SELECT date, mediatype FROM Email WHERE mediatype = 'Blog'

Any advice?

EDIT: Here is the table with:

There is more columns but these are ones I want to focus on.

Do you get an error? What happens when you run this directly in MySQL? — John Conde
– John Conde, Commented Jul 2, 2015 at 13:26
please show your mysql table setup. Also I'd recommend Using PDO and prepared statements. This has epic injection potential. — DerStoffel
– DerStoffel, Commented Jul 2, 2015 at 13:26
It does the same and returns nothing, there isn't any error. Maybe something wrong with the sql? — MrHappySandwich
– MrHappySandwich, Commented Jul 2, 2015 at 13:28
You generate an extra comma (,) before the FROM keyword and the query doesn't run at all because it has this syntax error. — axiac
– axiac, Commented Jul 2, 2015 at 13:30
What happens if you run SELECT date, mediatype FROM Email WHERE mediatype LIKE 'Blog%'? I'm thinking you have some trailing spaces or something in the data. — Samsquanch
– Samsquanch, Commented Jul 2, 2015 at 13:36

Thibault Henry · Accepted Answer · 2015-07-02 13:38:29Z

1

Your generate SQL request seems to have a syntax error. Just change the way you generate it.

Instead of

foreach($columns as $val)
  $query .= "$val, ";

Try

$query .= implode(', ' $columns);

That will skip the last comma.

answered Jul 2, 2015 at 13:38

Thibault Henry

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1 Comment

MrHappySandwich Over a year ago

Thanks, that is a much easier way than I was doing. However it still doesn't fix the problem.

MrHappySandwich · Accepted Answer · 2015-07-02 13:58:25Z

0

The blog column had an extra empty line that 'Blog%' wasn't working on. I went in the database and deleted the extra line and used the query again and it worked.

Thanks everyone for the help :)

answered Jul 2, 2015 at 13:58

MrHappySandwich

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