如何正确返回并显示json数据和getJSON？

Question

文件 “zip.json”： { “zip11111”： “City1”， “zip99999”： “城2” }

的JavaScript：

jQuery(document).ready(function() { 
$("#textbox1zip").change(function() { 
    $.getJSON('zip.json', function(data) { 

    /* each of the following code blocks don't work when exchanged for each other: 

    1 zipcode = "zip" + $("#textbox1zip").val(); 
    * $('#textbox2city').val(data.zipcode); -----> call silently dropped 

    2 zipcode = 'zip99999'; 
    * $('#textbox2city').val(data.zipcode); -----> call silently dropped 

    3 zipcode = "zip" + $("#textbox1zip").val(); 
    * alert(data.zipcode); -----> returns 'undefined' 
    */ 

    /* but each of these blocks here works: 
    4 $('#textbox2city').val(data.zip99999); ------> ok 

    5 alert(data.zip99999); -----> ok 

    6 zipcode = 'zip99999'; 
    * $('#textbox2city').val(zipcode); ------> ok 

    7 zipcode = "zip" + $("#textbox1zip").val(); 
    * $('#textbox2city').val(zipcode); ------> ok 
    */ 
    }); 
}); 
}); 

我想从获取的值textbox1zip将适当的值从“zip.json”返回到textbox2city。我想用getJSON。这也可能是很简单的，但我不能看到它... ...

Answer 1

试试这个：

var zipcode = 'zip99999'; // or 'zip11111' 
$('#textbox2city').val(data[zipcode]); 

和

var zipcode = "zip" + $("#textbox1zip").val(); 
$('#textbox2city').val(data[zipcode]); 

现在，这alert(data.zipcode); -----> returns 'undefined'

这样呢，因为JSON对象你只有两个键zip11111 & zip99999。所以，data.zip11111或data.zip99999的作品，而data.zipcode没有，因为没有名称zipcode的关键。