PHP MYSQL compare database value to POST value

Question

This is a really simple one, I just can't get my head around it sorry. I have this PHP code which picks up my form value, then compares it with the value stored in the database. That works fine.

However I am not sure how to write this logic in terms of this query:

If posted value = database value { // do something } else { // do something else }

if (empty($_POST['order_id']) === false) {
    // prepare data for inserting 
    $order_id = htmlentities(trim($_POST['order_id']));
    $order_id = preg_replace("/[^0-9]/","", $order_id);

    $result = mysqli_query($con,"SELECT * FROM listings WHERE order_id = $order_id");
    $row    = mysqli_fetch_assoc($result);

    echo $row['order_id'];

}

SOLVED:

Solved the question, was a silly one I know! Just needed this at the end of the code:

if($order_id === $row['order_id']) {
    echo 'found';
} else {
    echo 'not found';
}

Consider getting the number of results affected by your query and evaluate off of that value. If that number is 0, then we didn't find any matching results. You can then take the user down the else path. Otherwise, you can take them down the happy path. — James Binford
– James Binford, Commented Dec 31, 2013 at 18:36
does $_POST['order_id'] contain an id that contains only one non-digit character. The preg_replace function in your code does remove only one non-digit character. — Rajesh Paul
– Rajesh Paul, Commented Dec 31, 2013 at 19:00

Kei · Accepted Answer · 2013-12-31 18:37:17Z

3

Try

If ($result->num_rows === 1) { do something } else { do something else }

answered Dec 31, 2013 at 18:37

Kei

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2 Comments

Cobra_Fast Over a year ago

num_rows > 0 can also be considered, depending on the behaviour OP wants.

Kei Over a year ago

True, though he's launching mysqli_fetch_assoc only once, I assumed the result was expected to be only one.

David Nguyen · Accepted Answer · 2013-12-31 18:36:44Z

0

Since you did the business logic in your query you can just use

if( ! is_null($row)) {
   // do
} else {
   // nothing
}

answered Dec 31, 2013 at 18:36

David Nguyen

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Comments

Digital Chris · Accepted Answer · 2013-12-31 18:39:53Z

0

Did I read too much into "If posted value = database value "? Are you just referring to the order_id?

if ($row['listingName'] == $_POST['frm_listingName']) {
    // something
}
else {
    //something else
}

answered Dec 31, 2013 at 18:39

Digital Chris

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Comments

Rajesh Paul · Accepted Answer · 2013-12-31 18:55:04Z

0

Check this code:

if (empty($_POST['order_id']) === false) {
    // prepare data for inserting 
    $order_id = htmlentities(trim($_POST['order_id']));
    $order_id = preg_replace("/[^0-9]/","", $order_id);

    $result = mysqli_query($con,"SELECT * FROM listings WHERE order_id = $order_id");

    if(mysqli_num_rows($result)>0)
    {
        //Match found. do something.
    }
    else
    {
        //No match found. do something.
    }
}

N.B. In place of mysqli_num_rows($result) you can also use $result->num_rows

answered Dec 31, 2013 at 18:55

Rajesh Paul

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