您好,这是代码,因此,如果可以正常工作并向我显示消息,但其他方法不起作用,则不会显示任何消息尝试了很多事情,但是当我输入正确的登录名和密码时,总是一样的,如果输入不正确,则尝试也可以使它失效! =在那里
export default class App extends React.Component {
constructor(props) {
super(props);
this.state = {
lists: [
{
name: 'sampleList',
id: 15,
permission: {
canRead: true,
canWrite: false
}
}
]
}
}
invertCanRead(targetId) {
this.setState({
...this.state,
lists: this.state.lists.map(list => {
if (list.id === targetId) {
// that is the one we want to modify, so let's return
// modified version:
return {
...list,
permission: {
...list.permission,
canRead: !list.permission.canRead
}
}
} else {
// else return list as is without changes because
// it's not the one we want to modify, so just copy it:
return list;
}
})
})
}
render() {
return <div>
{JSON.stringify(this.state)};
<button onClick={() => this.invertCanRead(15)}></button>
</div>
}
}
答案 0 :(得分:0)
您的else语句(您有很多,并且没有明确指定哪个)永远不会运行。如果没有(用户;通过)命中,那么您的那一阵子就不会有任何迭代。
实际上,出于相同的原因,if(emailLog == email && passwordLog == password)
是多余的。
您的代码段:
while(query.next())
{
QString emailLog = query.value(1).toString();
QString passwordLog = query.value(4).toString();
if(emailLog == email && passwordLog == password){
QMessageBox::information(this,"SUCCESS","SUCCESS");
db.close();
}
else
{
ui->plstryagain->show();
db.close();
}
}
也许更好:
if (query.hasNext()) {
QMessageBox::information(this,"SUCCESS","SUCCESS");
db.close();
} else {
ui->plstryagain->show();
db.close();
}
P / S: 每当您认为基本语言功能不起作用时,很可能是您的错:P