我正在将mysqlf引入WebSQL(是的,我知道它已被弃用,而我遇到了麻烦...
我通过Js创建数据库,如下所示:
function initDb(){
try {
if (!window.openDatabase) {
alert('Databases are not supported in this browser.');
} else {
var db = getDb();
db.transaction(function (tx){
tx.executeSql("CREATE TABLE IF NOT EXISTS tipofiesta (id INTEGER AUTO_INCREMENT, honor TEXT, descripcion_en TEXT, descripcion_es TEXT, descripcion_eu TEXT, PRIMARY KEY (id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS localizacion (id INTEGER AUTO_INCREMENT, latitud TEXT, longitud TEXT, PRIMARY KEY (id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS claseacto (id INTEGER AUTO_INCREMENT, nombre_en TEXT, nombre_es TEXT, nombre_eu TEXT, descripcion_en TEXT, descripcion_es TEXT, descripcion_eu TEXT, CONSTRAINT id_pk PRIMARY KEY (id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS AdministradorFiestas (id INTEGER AUTO_INCREMENT, nombreUsuario TEXT, pass TEXT, email TEXT, verificado BOOLEAN DEFAULT FALSE, CONSTRAINT id_pk PRIMARY KEY (id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS provincia (id INTEGER AUTO_INCREMENT, nombreCas varchar(25) DEFAULT 'nombre de provincia', nombreEus varchar(25) DEFAULT 'probintzia izena', id_localizacion INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_localizacion) REFERENCES localizacion(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS pueblo (id INTEGER AUTO_INCREMENT, nombreCas TEXT, nombreEus TEXT, id_localizacion INTEGER, id_provincia INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_localizacion) REFERENCES localizacion(id), FOREIGN KEY (id_provincia) REFERENCES provincia(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS fiesta (id INTEGER AUTO_INCREMENT, nombre TEXT, descripcion_eu TEXT, descripcion_en TEXT, descripcion_es TEXT, id_pueblo INTEGER, id_tipofiesta INTEGER, id_fechafiestaanyo INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_pueblo) REFERENCES pueblo(id), FOREIGN KEY (id_tipofiesta) REFERENCES tipofiesta(id), FOREIGN KEY (id_fechafiestaanyo) REFERENCES fechafiestaanyo(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS fechafiestaanyo (id INTEGER AUTO_INCREMENT, fechaInicio date, fechaFin date, id_fiesta INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_fiesta) REFERENCES fiesta(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS diafiesta (id INTEGER AUTO_INCREMENT, fecha date, nombre TEXT, descripcion_en TEXT, descripcion_es TEXT, descripcion_eu TEXT, id_fechafiestaanyo INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_fechafiestaanyo) REFERENCES fechafiestaanyo(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS acto (id INTEGER AUTO_INCREMENT, nombre_en TEXT, nombre_es TEXT, nombre_eu TEXT, descripcion_en TEXT, descripcion_es TEXT, descripcion_eu TEXT, horaInicio time, horaFinAprox time, id_localizacion INTEGER, id_diafiesta INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_localizacion) REFERENCES localizacion(id), FOREIGN KEY (id_diafiesta) REFERENCES diafiesta(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS actoclases (id INTEGER AUTO_INCREMENT, id_acto INTEGER, id_claseacto INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_acto) REFERENCES acto(id), FOREIGN KEY (id_claseacto) REFERENCES claseacto(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS AdministradorFiestasPueblo (id INTEGER AUTO_INCREMENT, idAdministradorFiestas INTEGER, idPueblo INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (idAdministradorFiestas) REFERENCES AdministradorFiestas(id), FOREIGN KEY (idPueblo) REFERENCES pueblo(id))", []);
tx.executeSql("CREATE TABLE IF NOT EXISTS tipofiestadefiesta (id INTEGER AUTO_INCREMENT, id_tipofiesta INTEGER, id_fiesta INTEGER, CONSTRAINT id_pk PRIMARY KEY (id), FOREIGN KEY (id_tipofiesta) REFERENCES tipofiesta(id), FOREIGN KEY (id_fiesta) REFERENCES fiesta(id))", []);
});
}
return;
} catch(e) {
if (e == 2) {
// Version number mismatch.
console.log("Invalid database version.");
} else {
console.log("Unknown error "+e+".");
}
return;
}
}
最大的问题是,每个表的“ id”列都不能与auto_increment子句一起使用...我可以使其工作吗?
否则,我正在创建一个函数,该函数告诉我表中可用的以下ID(以便添加数据等):
function nextIdInTableAvailable(tableName){
try {
console.log("entramos");
var idNotAvailable = false;
var anId = 0;
console.log("entramos al while...");
var db = getDb();
while(idNotAvailable==false){
console.log("toca vuelta " + anId);
var currentIdUsed = false;
db.transaction(function (tx){
tx.executeSql('SELECT * FROM ' + tableName + ' WHERE id=? ORDER BY id ASC', [anId], function (tx, results) {
console.log("hay resultados en la vuelta " + anId);
currentIdUsed = true;
});
});
if(!currentIdUsed){
idNotAvailable = true;
console.log("vamos a salir en la vuelta " + anId);
}else{
anId = anId + 1;
console.log("seguimos en vuelta siguiente: " + anId);
}
}
console.log("El siguiente id disponible para la tabla '" + tableName + "' es: " + anId );
return anId;
} catch(e) {
if (e == 2) {
// Version number mismatch.
console.log("Invalid database version.");
} else {
console.log("Unknown error "+e+".");
}
return 0;
}
}
但是,还是一个问题:事务异步运行,因此不需要返回的值...
如果auto_increment无法正常工作,此功能有解决方案吗?
由于事务异步运行,因此无法以简单的方式返回值...
https://stackoverflow.com/a/12462907/828551 @apsillers有解决方案:回调。
我使用ionic 4和SQLite。我想创建一个自动增量ID,但是每次在该表上插入数据时收到一条错误消息时,我都会创建一个。
这是我的表创建代码
this.database.executeSql('CREATE TABLE IF NOT EXISTS switch_types (switch_types_id INT(11) PRIMARY KEY ASC,image_path VARCHAR(150),status TINYINT(1),name VARCHAR(150),updated_at TIMESTAMP DEFAULT CURRENT_TIMESTAMP)', [])
.then(() => {
console.log('switch_types table created successfully');
this.statusOfSwitch();
})
.catch(e => console.log('error on switch_types table creation' + JSON.stringify(e)));
这里是我的数据插入代码
addSwitchTypes(typeObj): Observable<any> {
const switchTypeData = [typeObj.image_path, typeObj.status, typeObj.name];
return new Observable((obs) => {
this.database.executeSql('INSERT INTO switch_types (image_path, status, name) VALUES (?, ?, ?)', switchTypeData)
.then(() => {
console.log('switch type added successfully');
obs.next(true);
obs.complete();
}).catch((error) => {
console.log('switch type not added successfully', error);
obs.error(error);
obs.complete();
});
});
}
我收到的错误消息
message: "sqlite3_step failure: NOT NULL constraint failed: switch_types.switch_types_id"
我已经尝试了几种方法,但是失败了。谁能帮我吗?我已经尝试了以下
this.database.executeSql('如果不存在则创建表switch_types(switch_types_id INT(11)PRIMARY KEY ASC,image_path VARCHAR(150).......)
对于'auto_increment',您可以尝试一下,它对我有用:
CREATE TABLE IF NOT EXISTS 'yourTableName' ('idPrimaryKey' INTEGER PRIMARY KEY ASC, 'column2' VARCHAR(45) NULL,'column3' TINYINT(1) NULL);