Question

当我使用$ .post进行ajax调用时，我想要设置一些基本设置，所以每次我想使用$ .post时我都不需要将它放在我的代码中。我这样做了：

$.ajaxSetup({
    dataType    :"json", // all requests should respond with json string by default
    type        : "POST", // all request should POST by default
    beforeSend  : function(){
        this.url = basepath+"include/ajax/"+this.url; // before making the request, change url location to ajax folder
    },
    error       : function(xhr, textStatus, errorThrown){
        displayMessage("error", "Request could not be completed: <pre>"+errorThrown+"</pre>", true);
    },
    success : function(event){
    console.log(event);
        if(event.responseJSON !== undefined && event.responseJSON.status !== undefined && event.responseJSON.status === true){
            if(event.responseJSON.output === undefined){
                console.log("something is wrong 1");
                displayMessage("error", "Property output was not found in response.");
            }
            else{
                event.responseJSON.status = false;
                console.log("something is wrong 2");
            }
        }
        else{
            console.log("something is wrong 3");
        }
    },
    done    : function(){
        console.log("done");
    },
    always  : function(){
        console.log("always");
    }
});

我像这样制作我的$ .post：

$("div#brandsView button.compose").click(function(e){
    $.post("brands.php",{method:"composeMessage",data:{brand:brandId}},function(data){
        console.log(data)
    })
});

现在我遇到了一个问题：我想在$ .ajaxSetup中更改“brands.php”的响应，然后再转到$ .post。正如您所看到的，我尝试过成功和完成的功能，但这些不会改变$ .post中的结果。有谁知道这是否可能，如果有的话，该如何做？

祝你好运，克里斯

Answer 1

$.post中的成功处理程序会覆盖$.ajaxSetup中的成功处理程序。您应该从ajaxSetup：

中提取成功处理程序定义

function handleSuccess(event){
    console.log(event);
        if(event.responseJSON !== undefined && event.responseJSON.status !== undefined && event.responseJSON.status === true){
            if(event.responseJSON.output === undefined){
                console.log("something is wrong 1");
                displayMessage("error", "Property output was not found in response.");
            }
            else{
                event.responseJSON.status = false;
                console.log("something is wrong 2");
            }
        }
        else{
            console.log("something is wrong 3");
        }
    }

然后让你的$.ajaxSetup像这样：

$.ajaxSetup({
    dataType    :"json", // all requests should respond with json string by default
    type        : "POST", // all request should POST by default
    beforeSend  : function(){
        this.url = basepath+"include/ajax/"+this.url; // before making the request, change url location to ajax folder
    },
    error       : function(xhr, textStatus, errorThrown){
        displayMessage("error", "Request could not be completed: <pre>"+errorThrown+"</pre>", true);
    },
    success : handleSuccess,
    done    : function(){
        console.log("done");
    },
    always  : function(){
        console.log("always");
    }
});

然后让你的$.post像这样：

$.post("brands.php",{method:"composeMessage",data:{brand:brandId}},function(data){
        handleSuccess(data);
        console.log(data);
    })

$ .ajaxSetup更改响应$ .post

1 个答案: