php – SQL错误：SQLSTATE [23000]：完整性约束违规：1452无法添加或更新子行：外键约束失败

2023年5月5日 290次阅读

我有一些
PHP和SQL的问题,
PHP对我来说是新的,我缺乏SQL.

我想在我的数据库的表中添加值：值可以从表单中获取,也可以从另一个表中获取(作为FK).

这是数据库：

create database AAA;
use AAA;

create table assure(
id_assure varchar(13) not null,
nom varchar(20),
adresse varchar(50), mdp varchar(60),
primary key(id_assure));

create table vehicule(
id_vehicule varchar(13) not null,
immatriculation varchar(7),
masse int(4),
volume int(4),
id_assure varchar(13) not null,
primary key(id_vehicule),
foreign key(id_assure) references assure(id_assure));

create table reparation(
id_reparation varchar(13) not null,
libelle varchar(100),
couts int(5),
primary key(id_reparation),
id_vehicule varchar(13) not null,
foreign key(id_vehicule) references vehicule(id_vehicule));


create table sinistre(
id_sinistre varchar(13) not null,
libelle varchar(100),
date_sinistre date,
heure_sinistre time,
primary key(id_sinistre),
id_vehicule varchar(13) not null,
foreign key(id_vehicule) references vehicule(id_vehicule));

create table dossier(
id_dossier varchar(13) not null,
primary key(id_dossier),
id_sinistre varchar(13) not null,
foreign key(id_sinistre) references sinistre(id_sinistre));

create table contrat_assurance(
id_assurance varchar(13) not null,
primary key(id_assurance),
id_assure varchar(13)  not null,
id_vehicule varchar(13) not null,
foreign key(id_assure) references assure(id_assure),
foreign key(id_vehicule) references vehicule(id_vehicule));

create table type_garantie(
code_garantie varchar(13) not null,
libelle varchar(100),
franchise int(3) ,
primary key(code_garantie),
id_assurance varchar(13) not null,
id_reparation varchar(13) not null,
foreign key(id_reparation) references reparation(id_reparation),
foreign key(id_assurance) references contrat_assurance(id_assurance));


DELIMITER //
CREATE TRIGGER trigfranchise
 BEFORE INSERT ON type_garantie
 for EACH ROW
BEGIN
    IF NEW.franchise not between 150 and 600

    THEN
    SET NEW.franchise = NULL ;
    END IF;
END ;//
DELIMITER ;

DELIMITER //
CREATE TRIGGER trigfranchise1
 BEFORE UPDATE ON type_garantie
 for EACH ROW
BEGIN
    IF NEW.franchise not between 150 and 600

    THEN
    SET NEW.franchise = OLD.franchise ;
    END IF;
END ;//
DELIMITER ;

我的第一个表单发送值以确保表,它工作正常.
我的第二个意思是将值发送到vehicule和contrat_assurance表;检查表单后,浏览器显示：

Erreur : SQLSTATE[23000]: Integrity constraint violation: 1452 Cannot
add or update a child row: a foreign key constraint fails
(aaa.vehicule, CONSTRAINT vehicule_ibfk_1 FOREIGN KEY
(id_assure) REFERENCES assure (id_assure))

我的PHP代码如下：

try
        {
            $nom=mysql_real_escape_string($_POST["nom"]);
            $mdp=mysql_real_escape_string($_POST["mdp"]); 
            $imma=mysql_real_escape_string($_POST["immatriculation"]); 

            $pdo_options[PDO::ATTR_ERRMODE] = PDO::ERRMODE_EXCEPTION;
            $bdd = new PDO('mysql:host=localhost;dbname=AAA', 'root', '',   $pdo_options);

            $req = $bdd->prepare('INSERT INTO vehicule (id_vehicule, immatriculation) VALUES(:id_vehicule, :immatriculation)');

            $req->execute(array(
            'id_vehicule' => uniqid(),
            'immatriculation' => $imma
            ));


            $req2 =  $bdd->prepare("INSERT INTO vehicule (id_assure)  SELECT id_assure FROM assure WHERE nom =\"$nom\" AND mdp =\"$mdp\"" );



            $req3 = $bdd->prepare('INSERT INTO contrat_assurance (id_assurance) VALUES(:id_assurance)');
            $req4 =  $bdd->prepare("INSERT INTO contrat_assurance (id_assure, id_vehicule)  SELECT id_assure, id_vehicule FROM vehicule WHERE immatriculation =\"$imma\"");

            $req3->execute(array(
            'id_assurance' => uniqid()

            ));



        }

实际上,我找到了关于这个错误的其他主题,由于我的技能很差,我无法将答案与我的问题相匹配.

对不起英文和法文表名不佳.

谢谢你的阅读！

更新：我试图通过终端(没有PHP)添加值,它工作,因此我的PHP代码是错误的.

UPDATE2：这是将id_assure插入到保证表(它是vehicule表中的FK)的PHP代码：

try
        {
            $pdo_options[PDO::ATTR_ERRMODE] = PDO::ERRMODE_EXCEPTION;
            $bdd = new PDO('mysql:host=localhost;dbname=AAA', 'root', '',   $pdo_options);

            $req = $bdd->prepare('INSERT INTO assure(id_assure, nom, adresse, mdp) VALUES(:id_assure, :nom, :adresse, :mdp)');
            $req->execute(array(
            'id_assure' => uniqid(),
            'nom' => $_POST['nom'],
            'adresse' => $_POST['adresse'],
            'mdp' => $_POST['mdp']
            ));


        }
        catch(Exception $e)
        {
            die('Erreur : '.$e->getMessage());
        }

最佳答案 问题是,当您插入到vehicule时,您需要指定外键,即您需要将值插入其中,并且该值需要存在于引用表中.在这种形式中,您插入的唯一列是id_vehicule和immatriculation,这意味着id_assure将保持为空,这意味着约束将失败.

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