着訪問JSON

Question

我有這樣的Ajax請求，

$.ajax({ 
    url : '<?php echo current_url(); ?>', 
    data  : "txtKeywords="+$("#txtKeywords").val()+"&search=Search For Item", 
    type  : 'POST', 
    dataType : 'JSON', 
    success : function(html) 
    {  
     console.log(html);   
    } 
}); 
return false; 

我得到了我的控制檯下面，

[{"productId":"5","productTitle":"Small Brasserie Dining Table","productPath":"small-brasserie-dining-table\/","productRangeId":"6","productSecondaryRangeId":"0","productTypeId":"2","productRefNo":"0080","productShortDesc":"","productBestSeller":"0","productFeatured":"0","productIsSet":"0","productPrice":"275","productSavingType":"none","productSavingPrice":"0","productSavingMessage":"","productDimWidth":"90","productDimHeight":"74","productDimDepth":"90","productTechnical":"Powder coated aluminium frame with welded joints.","productTemplateId":"5","productMetadataTitle":"","productMetadataKeywords":"","productMetadataDescription":"","productBandingColour":"grey","productActualPrice":"275","rangeTitle":"Dining","parentRangeTitle":"Aegean","fullRangePath":"aegean\/dining\/","fullProductPath":"aegean\/dining\/small-brasserie-dining-table\/","hasImage":"0"}] 

但是，當我這樣做，

alert(html.productTitle) 

所有我get是不確定的？

我在做什麼錯【J

Answer 1

嘗試HTML [0] .productTitle，我也碰到這個問題了幾次。

Answer 2

是因爲你的html變量是一個數組嗎？難道你要做......

alert(html[0].productTitle); 

Answer 3

嘗試使用jQuery的JSON AJAX功能，而不是$阿賈克斯將解析JSON你。

http://api.jquery.com/jQuery.getJSON/

Answer 4

嘗試做這樣的事情：

alert(html.d); // this will show the result of your ajax call

我不知道原因是使用屬性「d」什麼，但如果你可以看看你的結果，使用一個調試工具來查看哪些數據正在恢復您的ajax調用。

，如果你想你的JSON字符串轉換爲與目標，你可以用下面的代碼做：

var respuesta = jQuery.parseJSON(html.d);