以模态

Question

在我的网站上，我试图制作一个模式，告知用户他们的用户名（根据他们的姓名和姓氏生成），并通知用户他们已被添加到审核中，以便他们的帐户获得批准。我在一个用ajax调用的php文件中创建帐户，然后使用JSON反馈一个整数，通知javascript进程是否成功，如果是，用户的用户名是什么。如果创建了帐户并且显示了有关批准过程的信息，我的模式就会打开，但是我似乎无法显示带有用户名的部分。您能否尝试查看我出错的地方。< / p>

的Javascript

var RegisterSuccess = new Array();
var RegisterSuccess = JSON.parse(Feedback);
if (RegisterSuccess[0] == 0){
    $('#mdlInfo').html('<p>Your account has been created under the username: "<strong><span id="spnUsername"></span></strong>". You <strong>must</strong> remember this as you will require it to login to your account.</p>');
    $('#mdlInfo').html('<p>Your account has also been added to a moderation que. You must wait until a member of staff activates your account!</p>');
    $('#spnUsername').html(RegisterSuccess[1]);
    $("#mdlRegister").modal("show");

PHP输出代码

$ Feedback = json_encode（array（$ RegisterSuccess，$ username））;

echo $ Feedback;

提前致谢！

Answer 1

为什么不从PHP返回正确的json_encoded数组？无需分配变量，只需这样做 echo json_encode(["success"=>"true","username"=>"$username"]); 并且在javascript中（不需要将RegisterSuccess指定为数组），就这样做 var returnValue = JSON.parse(Feedback); （你应该避免调用变量RegisterSuccess - 选择一个中性名称，这样你就可以将它用于成功和错误而没有问题。然后就这样做了

$('#mdlInfo').html('<p>Your account has been created under the username: <strong><span id="spnUsername">'+returnValue['username']+'</span></strong>. You <strong>must</strong> remember this as you will require it to login to your account.</p><p>Your account has also been added to a moderation que. You must wait until a member of staff activates your account!</p>');
$("#mdlRegister").modal("show");