在带有JSON对象的jQuery中使用.each时，如何避免未定义的属性？

Question

I'm attempting to fill the values of <inputs> on a page using the data from a JSON object.

http://codepen.io/jimmykup/pen/exAip?editors=101

First I create my JSON object. In this case it only has two values. One name and one url .

var jsonObj = [];

var name = "1stname";
var url = "firsturl";

item = {}
item ["name"] = name;
item ["url"] = url;

jsonObj.push(item);

Next I use jQuery's .each() to find all inputs with a class of .name and I use the index to insert the value of "name" into the input. After that I want to go through all of the inputs with .url as the class name and do the same.

$("input.name").each(function( index ) {
    if ( jsonObj[index].name ) {
      $(this).val( jsonObj[index].name );
  }
});

$("input.url").each(function( index ) {
    if ( jsonObj[index].url ) {
      $(this).val( jsonObj[index].url );
  }
});

The problems is that if my JSON object has 1 value for "name" but .each() finds 2 inputs, once it gets to the second input I get this error in my console.

Uncaught TypeError: Cannot read property 'name' of undefined

Then my second .each() function fails to run. What can I do to avoid this issue?

Answer 1

You need to make sure jsonObj[index] is defined before trying to get the name or url .

$("input.name").each(function( index ) {
  var data = jsonObj[index];
  if ( data && data.name ) {
    $(this).val( data.name );
  }
});

$("input.url").each(function( index ) {
  var data = jsonObj[index];
  if ( data && data.url ) {
    $(this).val( data.url );
  }
});

Answer 2

Json objet is in this form::

var jsonobj={"name":"1stname","url":"firsturl"};

with this format you can execute your code.

and what you are forming is basically an array.

Answer 3

Check that the index is less than the length of the array, otherwise quit the loop:

$("input.name").each(function( index ) {
  if (index < jsonObj.length && jsonObj[index].name ) {
    $(this).val( jsonObj[index].name );
  } else {
    return false;
  }
});

$("input.url").each(function( index ) {
  if ( index < jsonObj.length && jsonObj[index].url ) {
    $(this).val( jsonObj[index].name );
  } else {
    return false;
  }
});

Fork