Output from imploded array not returning first instance (PHP)

Question

So, I have an array which I am combining to create an INSERT command. (No sanitizing done yet, just a proof of concept).

$namedata is the data for the "Name" field (for example "This Name") $refdata is the data for the "Ref" field (for example "43857368") $CatRefData is my array (for example, '23', '45', '2', '144')

I am formulating it as follows:

$inputdata = "'), ('".$namedata."', '".$refdata."', '";
  $data = implode($inputdata, $CatRefData);
  $result = "INSERT INTO `MyTable`('Name', 'Ref', 'Cat_Ref') VALUES ('".$data."')";

But the output comes out with the first instance of the array missing any of the additional "non-array" data, like this:

$result = INSERT INTO MyTable('Name', 'Ref', 'Cat_Ref') VALUES ('23'), ('This Name', '43857368', '45'), ('This Name', '43857368', '2'), ('This Name', '43857368', '144')

Can you see - the first instance, where the array value is '23', doesn't have any of the surrounding information.

What am I doing wrong?

i think this doesnt work with implode, so you have do do a foreach to fill your $data. — steven
– steven, Commented Jul 9, 2013 at 14:35

steven · Accepted Answer · 2013-07-09 14:38:03Z

1

I would do it like this...

$arr = array();

foreach($CatRefData as $d) {
    $arr[] = "('".$namedata."', '".$refdata."', '".$d."')";

}

$data = implode(",",$arr);

but you should do some escaping...

answered Jul 9, 2013 at 14:38

steven

4,8952 gold badges31 silver badges61 bronze badges

Sign up to request clarification or add additional context in comments.

1 Comment

user1259798 Over a year ago

Ah, "foreach"! Thanks - a great solution. Perfect! (I will be escaping each of the values before they get as far as this formula, but thank you for the advice all the same).

Marc B · Accepted Answer · 2013-07-09 14:38:44Z

1

implode() only adds the join text BETWEEN values, so if you have:

$arr = array('a', 'b', 'c');
$newarr = implode('foo', $arr);

you'll end up with

afoobfooc

but you actually need

fooafoobfooc

That means you have to "prime" your query with the "join" text first:

$result = "INSERT INTO `MyTable`('Name', 'Ref', 'Cat_Ref') VALUES ('$inputData"

answered Jul 9, 2013 at 14:38

Marc B

362k44 gold badges433 silver badges508 bronze badges

Comments

Karo · Accepted Answer · 2013-07-09 14:38:27Z

0

I don't think you can do this with implode only.

Try using it with array_map() , something like this:

$CatRefData = [ '23', '45', '2', '144'];
function surrounding($n)
{
     return "('something', '123', '".$n."')";
}
echo implode(", ",array_map("surrounding", $CatRefData));

answered Jul 9, 2013 at 14:38

Karo

7444 silver badges7 bronze badges

Collectives™ on Stack Overflow

Output from imploded array not returning first instance (PHP)

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

3 Answers 3

1 Comment

Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related