variable in SQL INSERT

Question

I'm new in SQL, I would like to know how I can have this code working. Because it doesn't understand that $lala and $lolo are variable.

$lolo = "ahha";
$lala = "2";
$sql = "INSERT INTO organisation (id_orga, nom) VALUES ";
$sqll = '($lala, $lolo)';

$resultat = $conn->query($sql.$sqll);

so I tried:

$lolo = "ahha";
$lala = "2";
$sql = "INSERT INTO organisation (id_orga, nom) VALUES ";
$sqll = '('.$lala.','. $lolo.')';

$resultat = $conn->query($sql.$sqll);

But both code don't INSERT in my BDD and also give no error.

even at the start, always use prepared statements: php.net/manual/de/mysqli.quickstart.prepared-statements.php — neurotic-d
– neurotic-d, Commented Jun 16, 2016 at 11:49
Also which DBMS are you using? And what programming language is that? — user330315
– user330315, Commented Jun 16, 2016 at 11:56

M.M · Accepted Answer · 2016-06-16 11:57:02Z

1

In PHP the varibles are declared like so:

$lala = "ahah";

When you need to echo or print them you can do the string concatenation like echo 'this is my' . $lala; OR echo "this is my $lala";

This happens because the double quotes allow you to insert the variable without the need to concatenate. So, in your case:

$sqll = '('.$lala.','. $lolo.')';

is correct and is the same as:

$sqll = "($lala, $lolo)";

With the single quotes $sqll = '($lala, $lolo)'; wouldn't work as it would output ($lala, $lolo) and not (2, ahaha)

Note: take a look at pdo which is a better alternative to mysql security-wise

answered Jun 16, 2016 at 11:57

M.M

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3 Comments

moskitos Over a year ago

Well thanks for your great answer but i figure it out here is the working code:

$lolo = "ahha";         $lala = "2";         $sql = "INSERT INTO organisation (id_orga, nom) VALUES ";         $sqll = "('".$lala."','". $lolo."')";          $resultat = $conn->query($sql.$sqll);         echo $sql.$sqll;

Apostolos Over a year ago

yes you also need the single quotes for inserting strings in db and writing them in your insert statement

Apostolos Over a year ago

please also select an answer and upvote the ones that helped you. thnx

Apostolos · Accepted Answer · 2016-06-16 11:49:47Z

1

try with double quotes

$sqll = "($lala, $lolo)";

answered Jun 16, 2016 at 11:49

Apostolos

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1 Comment

Apostolos Over a year ago

see also stackoverflow.com/questions/5605965/…

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