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HomeBackend DevelopmentPHP Tutorialajax 从一个页面传递错误信息至另一个页面?怎么解决?

这是一个页面的代码 index.php

<code>   <script>
    $("#yollash").click(function() {    
        var isim=$("#uyisim").val();
        var mezmun=$("#qay_num").val();
        if( isim.length>=2 & mezmun.length>0 ){            
            var formData = $("#jediwel").serializeArray();
            $.ajax({
                url : "yollash.php",
                type: "POST",
                data : formData,
                success: function(data, textStatus, jqXHR)
                {
                    $(".modal-body").html("<h2>قۇشۇلدى!")
                    $('#myModal').modal('toggle')
                 },
                 error: function(jqXHR, textStatus, errorThrown)
                {
                    $(".modal-body").html("<h2>قايتۇرۇش نۇمۇرى بار ئىكەن!")
                    $('#myModal').modal('toggle')
                }
            });
        } else {
            $(".modal-body").html("<h2>تۇلۇق تولدۇرۇڭ")
            $('#myModal').modal('toggle')
        }
    });
 </script>
</code>

我做了另一个页面,就是把表单上的数据插入到数据库中,然后就在那个页面判断数据库中某个数据存在不存在,如果存在就会传递给前面的页面一个错误信息,下面的是另一个页面的代码片段 yollash.php

<code>$yengi = mysql_query("SELECT qay_num FROM wx_milodiye WHERE qay_num= '$qay_num'"); 
if(mysql_num_rows($yengi)) {  

} else { 
    $yollash_sql= mysql_query("INSERT INTO wx_milodiye(uy_ismi,tur,musicurl,hqmusicurl,qay_num,waqit) VALUES('$uyisim','$tur','$musicurl','$hqmusicurl','$qay_num',now())") or die (mysql_error());   
}
</code>

就在那个判断的地方怎么写?(新学者,可能表达不清楚)

回复内容:

这是一个页面的代码 index.php

<code>   <script>
    $("#yollash").click(function() {    
        var isim=$("#uyisim").val();
        var mezmun=$("#qay_num").val();
        if( isim.length>=2 & mezmun.length>0 ){            
            var formData = $("#jediwel").serializeArray();
            $.ajax({
                url : "yollash.php",
                type: "POST",
                data : formData,
                success: function(data, textStatus, jqXHR)
                {
                    $(".modal-body").html("<h2>قۇشۇلدى!")
                    $('#myModal').modal('toggle')
                 },
                 error: function(jqXHR, textStatus, errorThrown)
                {
                    $(".modal-body").html("<h2>قايتۇرۇش نۇمۇرى بار ئىكەن!")
                    $('#myModal').modal('toggle')
                }
            });
        } else {
            $(".modal-body").html("<h2>تۇلۇق تولدۇرۇڭ")
            $('#myModal').modal('toggle')
        }
    });
 </script>
</code>

我做了另一个页面,就是把表单上的数据插入到数据库中,然后就在那个页面判断数据库中某个数据存在不存在,如果存在就会传递给前面的页面一个错误信息,下面的是另一个页面的代码片段 yollash.php

<code>$yengi = mysql_query("SELECT qay_num FROM wx_milodiye WHERE qay_num= '$qay_num'"); 
if(mysql_num_rows($yengi)) {  

} else { 
    $yollash_sql= mysql_query("INSERT INTO wx_milodiye(uy_ismi,tur,musicurl,hqmusicurl,qay_num,waqit) VALUES('$uyisim','$tur','$musicurl','$hqmusicurl','$qay_num',now())") or die (mysql_error());   
}
</code>

就在那个判断的地方怎么写?(新学者,可能表达不清楚)

<code>echo json_encode(array('code'=>201,'message'=>'数据已存在'));
</code>

js里面的success回调函数里面就可以

<code>if(data.code == 201){
    alert(data.message);
}
</code>

yollash.php 页面直接输出需要的内容就 ok 了

其实你需要定的是前后端交互的一个数据格式,到底是文本内容还是json数据。

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