javascript - JavaScript 调用约定有问题

Question

这是代码的快照:

如果 Line:3 处的函数 testFields 返回 false，则控件将正确转到 Line:21 并返回 false.如果 testFields 返回 true，则控件将转到 Line:4，然后不再继续处理 函数中返回的响应Line:5，控件继续到 Line:21，其中 retVal 仍为 false。也就是说，在函数向函数返回值后检查 Line:6 的条件:

<form method="POST" onsubmit="return validateInput()" />

从逻辑上讲，它应该首先测试 Line:7 处的条件，然后决定 retVal 应该为 true 还是让它为真默认 false，然后应通过 Line:21 返回值。

我的代码在逻辑上有什么问题？

Answer 1

有两种方法。

1. 您可以发出同步ajax请求。

   var retVal ;
   $.ajax({ 
       url: youUrl,
       async: false, // required to pause script until request is done
       success: function( ret ) { 
          retVal = ret; // ret is invisible outside this callback
   
       },
       error : function( err ) {
         // your error logic
       }
   
   });
   //We'll in this line if request is completed and either success or error callback  was called.  
   return retVal;
   

2. 但是如果您不希望脚本被停止。您可以阻止表单中的默认操作，并从回调发送表单，例如:

   $( "form" ).submit( function( e, extra ) { 
   
     if( !extra )
       $.post( 
         url: "script.php",
         data: "yourdata",
         success: function( ) {  
           $( "form" ).trigger( "submit", [true] );         
         },
        error: function(){}
    );
   
      // extra parameter will be always undefined until we'll not pass it in an array which will be aplied to listener as  next ones arguments - take a look at success callback
     return !!extra  // doubled negate operator ensures that extra will be returned as boolean because if you return undefined  that means you didn't return anything 
   });
   

以及没有 onsubmit 声明的 html 表单标签:

<form method="POST" />