Implode() to make an array to string not working?

Question

I am stuck with this for hours, trying to fix it, but my little knowledge of php is not being enough, so I am asking for your help.

HTML

        <input type="checkbox" name="type[]" value="type1">Type 1<br>
        <input type="checkbox" name="type[]" value="type2">Type 2<br>
        <input type="checkbox" name="type[]" value="type3">Type 3<br>
        <input type="checkbox" name="type[]" value="type4">Type 4

PHP validate (separate file)

if (empty($_POST["type"])) {
    $typeErr = "This field is required";
} else {
    $type = array($_POST["type"]);
}

PHP

/*this is line 90*/ $type_string = implode(',', $type);

$sql = "INSERT INTO `table1`(`type`) VALUES ('$type_string')";

Error

 Notice: Array to string conversion in C:\xampp\htdocs\project\example.php on line 90

It keeps telling me I am converting an array to string when I use implode, I have no idea what I am doing wrong, I just have several checkboxes, and the selected ones go to an array, then I implode that array to a string to be able to send it to the mysql database...

It is probably a stupid mistake, but I am fairly new to php and mysql, so... can anyone help me?

I wrote $type_string to $typestring, but the problem remains, the notice still happens...

what @JohnConde says $typestring should be $type_string or the other way around. — Déjà vu
– Déjà vu, Commented Jun 23, 2014 at 14:39

exussum · Accepted Answer · 2014-06-23 14:42:19Z

1

$type = array($_POST["type"]);

now contains an array with 0 => array

So the implode is still trying to convert key 0 to a string.

Either

$type = (array) $_POST["type"];

Which is what i think you meant or

 /*this is line 90*/ $type_string = implode(',', $type[0]);

should fix it.

also make sure you defend against SQL injection

answered Jun 23, 2014 at 14:42

exussum

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2 Comments

NepsName Over a year ago

FINALLY, so thank you really, such a simple fix to something I was stuck for hours! Literally hours, I should have asked this here earlier... Oh well, thank you a lot, really, completely fixed it :)

NepsName Over a year ago

I am going to accept this answer as soon as it allows me (gotta wait 5min)

Rafal Kozlowski · Accepted Answer · 2014-06-23 15:03:07Z

0

$POST["type"] is already array. In Your code You are creating array(array()) srtucture. If You want to be sure You are working with array use

if(is_array($_POST["type"])){ $type=$_POST["type"] }

or cast to array by using

$type = (array)$_POST["type"];

then You are free to use implode Good luck :)

answered Jun 23, 2014 at 15:03

Rafal Kozlowski

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