dongnue4923 2013-06-18 21:07
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无法将PHP数组传递给jquery:JSON数据后出现意外的非空白字符

Here is what I have in PHP:

$arrayresults = array();
while($popularbrushesrow = mysql_fetch_array($popularBrushes)){

$arrayresults[]  = '<a href="brushdescription.php?id='.$popularbrushesrow['bd_brushid'].'"><img class="slideImg" alt="'.$popularbrushesrow['bd_brushname'].'" title="'. $popularbrushesrow['bd_brushname'].'" src="'.$popularbrushesrow['bd_imagefilepath'].'"  /></a>';

}

echo json_encode($arrayresults);

Now, jquery:

$.ajax({
    type:'GET',
    url:'getDataForSlide.php',
    data:"limit="+limit+"&required="+required,
    dataType:"json",
    cache:true,
        success: function(result){
     var arrayFromPHP = JSON.parse(result);
     alert(arrayFromPHP);
  }
})

Could someone please help me out. Whats the correct way to form an array in JSON?

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  • duanping1632 2013-06-18 21:11
    关注

    The problem is likely to be this line:

    var arrayFromPHP = JSON.parse(result);
    

    Because you've specified dataType: 'json' in the ajax options, jQuery has already done the parsing for you. So doing it a second time starts out by doing toString on the array, which does a join, which results in invalid JSON.

    Simply use result directly.

    For example, suppose you have this JSON:

    [
        "<a href=\"http://stackoverflow.com\">Stack Overflow</a>",
        "<a href=\"http://google.com\">Google</a>"
    ]
    

    Because jQuery has already done JSON.parse on it, result is an actual array. So if you then pass it into JSON.parse, the first thing that does is toString, which gives you this:

    <a href="http://stackoverflow.com">Stack Overflow</a>,<a href="http://google.com">Google</a>
    

    ...which is not, of course, valid JSON.

    本回答被题主选为最佳回答 , 对您是否有帮助呢?
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