iPhone中的sqlite

Question

我正在尝试在iPhone的phonegap中使用SQLite，我面临以下问题，请看一下下面的javascript代码。 非常感谢，

function testAsSeperateFunctionDB(contact)
{
  database.transaction(
    function( transaction ){
      alert("enteringIntoTransaction");
      transaction.executeSql(
        'INSERT INTO testTableThree (contactId, mrMrs, firstName, 
           lastName, jobTitle, bActive, homePhone, contactType, accountId, 
           workPhone, cellularPhone, workFax, contactEmail, owner, 
           alternateCountry, alternateAddress1, alternateAddress2, 
           alternateZipCode, alternateCity, alternateProvince) 
         VALUES (?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?, ?);',

        [contact.contactId, contact.mrMrs,
        contact.firstName, contact.lastName,
        contact.jobTitle, contact.bActive,
        contact.homePhone, contact.contactType,
        contact.accountId, contact.workPhone,
        contact.cellularPhone, contact.workFax,
        contact.contactEmail, contact.owner,
        contact.alternateCountry, contact.alternateAddress1,
        contact.alternateAddress2, contact.alternateZipCode,
        contact.alternateCity, contact.alternateProvince],

        //Callback after successful transaction
        function( transaction, results ) {
          alert("added");
        },

        //Callback after error in transaction 
        function( transaction, error ) {
          alert("Error" + error.code);
        }
      );
    }
  );
}

我正在使用上面的函数将一组联系人详细信息添加到表中，我将向此函数传递一个 JSON 对象。我在以下循环结构中调用上述函数。

for(i = 0; i < contactFullObjArray.length; i++) {
  //Parse the ContactFull object
  var contact=ContactFull.parse(contactFullObjArray[i]);    
  testAsSeperateFunctionDB(contact);        
}

这里发生的情况是，事务仅在整个 for 循环执行后才发生，并且仅被调用一次，这使得仅传递的最终值被添加到表中。 （我只收到一次警报“enteringIntoTransaction”，并且该值正在添加到数据库中。非常感谢您的帮助

Answer 1

我通过将

transaction.executeSql()

放在for 循环中并迭代它来解决这个问题。