檢查表存在

Question

我用這SELECT name FROM sqlite_master WHERE type = "table" AND name = 'create_organization';，以檢查是否存在名爲create_organization表，它works.However我想使用Qt發現out.I我試圖

QSqlQuery query; 
    QString tableName = "employee_datastores"; 
    QString sqlQuery = QString("SELECT name FROM sqlite_master WHERE type =:table AND name = %1 ").arg(tableName); 
    query.prepare(sqlQuery); 
    query.bindValue(":table", "table"); 
    query.exec(); 
    int fieldNo = query.record().indexOf("employee_datastore_name"); 
    while (query.next()) { 
     QString employee_ds_name = query.value(fieldNo).toString(); 
     qDebug() << "Table Name" << employee_ds_name ; 
    } 

，但它確實不行。

Answer 1

QSqlDatabase :: tables（）給你你想要的東西嗎？例如，你可以嘗試：

if (database.tables().contains(QLatin1String("employee_datastores")) { 
    ... 
} 

Answer 2

您可以修復它是這樣的：

QString sqlQuery = QString("SELECT name FROM sqlite_master WHERE type =:table AND name = '%1' ").arg(tableName); 
query.prepare(sqlQuery); 
query.bindValue(":table", "table"); 
query.exec(); 

但爲什麼混合綁定值和「啞巴」的字符串替換？

QString sqlQuery = QString("SELECT name FROM sqlite_master WHERE type =:table AND name = :tablename "); 
query.prepare(sqlQuery); 
query.bindValue(":table", "table"); 
query.bindValue(":tablename", tableName); 
query.exec(); 

Answer 3

這解決了它。

QSqlQuery query; 
    QString tableName = "structure"; 
    QString sqlQuery = QString("SELECT name FROM sqlite_master WHERE type =:table AND name = '%1' ").arg(tableName); 
    query.prepare(sqlQuery); 
    query.bindValue(":table", "table"); 
    query.exec(); 
    int fieldNo = query.record().indexOf("name"); 
    while (query.next()) { 
     QString _name = query.value(fieldNo).toString(); 
     qDebug() << "Table Name" << _name ; 
    }