外键约束的错误失败

Question

i'm creating a form to insert data to logement table and lit table and also table espace. i'm trying to insert data in all this tables that are connected like in the image enter image description here

but i'm getting this error :

Warning: mysql_insert_id() expects parameter 1 to be resource, object given in F:\-- SOFTWARE\WEB-SERVER\esyphp\ep0002\data\localweb\school\buckup\BU-Projet\0008\ajoutez.php on line 88
ERROR: Could not able to execute INSERT INTO logement(Titre, AdresseLogement, Prix, NombrePerson, idTypeLogement) VALUES ('hgfrte','125669 hgfer','1236','33',''). Cannot add or update a child row: a foreign key constraint fails (`yandexd`.`logement`, CONSTRAINT `FK_Logement_idTypeLogement` FOREIGN KEY (`idTypeLogement`) REFERENCES `typedelogement` (`idTypeLogement`)) ERROR: Could not able to execute INSERT INTO lit(idLit, TypeDeLit) VALUES ('','kanape, canape'). Cannot add or update a child row: a foreign key constraint fails (`yandexd`.`lit`, CONSTRAINT `FK_Lit_idLogement` FOREIGN KEY (`idLogement`) REFERENCES `logement` (`idLogement`)) 

my php code is

 if(isset($_POST['Submit'])){ /* Attempt MySQL server connection. Assuming you are running MySQL server with default setting (user 'root' with no password) */ $link = @mysqli_connect("localhost", "root", "", "yandexd"); // Check connection if($link === false){ die("ERROR: Could not connect. " . mysqli_connect_error()); } switch ($_POST['logm-type']) { case "Maison": $_POST['logm-type'] = 1; break; case "Appartement": $_POST['logm-type'] = 2; break; case "Chambre_privée": $_POST['logm-type'] = 3; break; case "Chambre_partagée": $_POST['logm-type'] = 4; break; } $sql = "INSERT INTO logement(Titre, AdresseLogement, Prix, NombrePerson, idTypeLogement) VALUES ('".$_POST['logm-titre']."','".$_POST['logm-adresse']."','".$_POST['logm-prix']."','".$_POST['logm-personne']."','".$_POST['logm-type']."')"; $thelogm_id = mysql_insert_id( $link ); $sql1 = "INSERT INTO espace(TypeDEspace, idLogement) VALUES ('".$_POST['logm-espace']."','".$thelogm_id."')"; $sql2 = "INSERT INTO lit(TypeDeLit, idLogement) VALUES ('".$_POST['logm-typelit']."','".$thelogm_id."')"; if(mysqli_query($link, $sql)){ echo "Le logement est ajouté."; } else{ echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); } if(mysqli_query($link, $sql1)){ echo "Le logement est ajouté."; } else{ echo "ERROR: Could not able to execute $sql1. " . mysqli_error($link); } if(mysqli_query($link, $sql2)){ echo "Le logement est ajouté."; } else{ echo "ERROR: Could not able to execute $sql2. " . mysqli_error($link); } // close connection mysqli_close($link); } else { 

I tried all solution found on stackoverflow, but i can't solve the problem

Answer 1

If the client sends you a string in logm-type that doesn't exactly match one of the four you've mentioned in your switch statement, your code will pass the string they gave you as if it was the ID, in which case the error message you're seeing is the best case scenario.

Your code is using POST values in SQL queries directly, which is not a good idea - what if the client (accidentally or deliberately) includes an apostrophe?

Answer 2

now i changed the code to

 $sql = "INSERT INTO logement(Titre, AdresseLogement, Prix, NombrePerson, idTypeLogement) VALUES ('".$_POST['logm-titre']."','".$_POST['logm-adresse']."','".$_POST['logm-prix']."','".$_POST['logm-personne']."','".$_POST['logm-type']."')"; $thelogm_id = mysql_insert_id( $link ); $sql1 = "INSERT INTO espace(TypeDEspace, idLogement) VALUES ('".$_POST['logm-espace']."','".$thelogm_id."')"; $sql2 = "INSERT INTO lit(TypeDeLit, idLogement) VALUES ('".$_POST['logm-typelit']."','".$thelogm_id."')"; 

but i did received this error

Warning: mysql_insert_id() expects parameter 1 to be resource, object given in F:-- SOFTWARE\\WEB-SERVER\\esyphp\\ep0002\\data\\localweb\\school\\buckup\\BU-Projet\\0008\\ajoutez.php on line 87 Le logement est ajouté.Le logement est ajouté.ERROR: Could not able to execute INSERT INTO lit(TypeDeLit, idLogement) VALUES ('sofa, canape, lit',''). Cannot add or update a child row: a foreign key constraint fails ( yandexd . lit , CONSTRAINT FK_Lit_idLogement FOREIGN KEY ( idLogement ) REFERENCES logement ( idLogement ))

Answer 3

Your calling mysql_insert_id out of sequence. It returns the generated ID from the last successful query, which means you have to execute $sql first.

Try this:

// this declares the sql query, but hasn't executed it. 
$sql = "INSERT INTO logement(Titre, AdresseLogement, Prix, NombrePerson, idTypeLogement)
    VALUES ('".$_POST['logm-titre']."','".$_POST['logm-adresse']."','".$_POST['logm-prix']."','".$_POST['logm-personne']."','".$_POST['logm-type']."')";

// this executes the sql.  Now mysql_insert_id can fetch the id. 
if(mysql_query($link, $sql)){
    echo "Le logement est ajouté.";
} else{
    echo "ERROR: Could not able to execute $sql. " . mysqli_error($link);
}

// call mysql_insert_id after executing $sql, otherwise you'll just get null
$thelogm_id = mysql_insert_id( $link );
if( is_null($thelogm_id)){
    echo "ERROR: no id".
}

$sql1 = "INSERT INTO espace(TypeDEspace, idLogement)
    VALUES ('".$_POST['logm-espace']."','".$thelogm_id."')";


$sql2 = "INSERT INTO lit(TypeDeLit, idLogement)
    VALUES ('".$_POST['logm-typelit']."','".$thelogm_id."')";  

if(mysqli_query($link, $sql1)){
    echo "Le logement est ajouté.";
} else{
    echo "ERROR: Could not able to execute $sql1. " . mysqli_error($link);
}

Be aware, if the query for $sql fails you'll not want to execute $sql1 and $sql2.