【问题标题】:How to retrieve data that is in a table? PostgreSQL, AngularJS1, Spring, Hibernate如何检索表中的数据? PostgreSQL、AngularJS1、Spring、休眠
【发布时间】:2018-05-07 08:15:36
【问题描述】:

在我的数据库中,我有姓氏和名字。我将它们收集在一个数组中;我的变量 lastAndFirstNameList 包含:例如:[name, first name] => ( ["WOOD", "Robin"])

然后我尝试使用变量 lastname 和 firstname 分别检索姓氏和名字。我得到结果 « lastname undefined » 和 « firstname undefined »。我不懂为什么!请问你能帮帮我吗?

enter code here : //controller.js
            Employee.getFirstAndLastName().$promise.then(function(result) {
                var lastAndFirstNameList = result.list; 
                var lastname = lastAndFirstNameList.lastName;
                var firstname = lastAndFirstNameList.firstName;

                for(var k = 0; k < lastAndFirstNameList.length; k++)
                    {
                        console.log("lastname", lastname);
                        console.log("firstname", firstname);
                    }


                console.log("lastAndFirstNameList", lastAndFirstNameList);
        }

....

enter code here daoImpl.java
@SuppressWarnings("unchecked")
@Override
public List<Object> getFirstAndLastName() {
    SQLQuery querySQL = sessionFactory.getCurrentSession().createSQLQuery("select last_name, first_name from employee");
    List<Object> firstAndLastNameList = querySQL.list();
    return firstAndLastNameList;
}

【问题讨论】:

    标签: postgresql hibernate spring-mvc


    【解决方案1】:

    这给了你什么?我认为 'firstname' 和 'lastname' 不会给你任何东西,因为它们没有被正确引用。

    Employee.getFirstAndLastName().$promise.then(function(result) {
                var lastAndFirstNameList = result.list; 
                var lastname = lastAndFirstNameList.lastName;
                var firstname = lastAndFirstNameList.firstName;
    
                console.log("lastname ", lastname);
                console.log("firstname ", firstname);
                console.log("lastAndFirstNameList", lastAndFirstNameList);
    }
    

    把上面的引用变量改成这样:

    var lastname = lastAndFirstNameList[0].lastName;
    var firstname = lastAndFirstNameList[0].firstName;
    

    如果从服务器正确传递,那应该会给你第一个值。

    您的循环应该如下所示:

    for(var k = 0; k < lastAndFirstNameList.length; k++){
        console.log("lastname", lastAndFirstNameList[k].lastName);
        console.log("firstname", firstname[k].firstName);
    }
    

    【讨论】:

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