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當我單擊按鈕時,我有上面的JSON錯誤。它不返回任何值。可以請你幫我解決這個問題,parsererror SyntaxError:JSON.parse:JSON數據第1行JSON數據後的JSON數據後的意外非空白字符
這裏我jQuery代碼,
$(document).ready(function(){
$('.package_details').click(function(){
$('.txt-form').slideDown("slow");
var package = $(this).attr("id");
$.ajax({
type:"post",
url:"aaa.php",
dataType: 'json',
data:"pack="+pack,
success:function(html){
$('.package_name').val(html.package_name);
$('.package_price').val(html.package_price);
$('.discount_percentage').val(html.discount);
$('.amount_after_discount').val(html.amnt_discount);
$('.tax_percentage').val(html.tax);
$('.amount_after_tax').val(html.amnt_tax);
if(html.package_type == 'Contact Visible'){
$('.contact').attr('checked', 'checked');
}else if(html.package_type == 'Job post'){
$('.job').attr('checked', 'checked');
}else if(html.package_type == 'Both'){
$('.both').attr('checked', 'checked');
}
$('.contact_visible').val(html.count);
$('.job_post').val(html.job_post);
$('.valitity_day').val(html.valitity);
},
error: function (jqXHR, textStatus, errorThrown) {
alert('Not done - ' + textStatus + ' ' + errorThrown);
}
})
})
})
,在這裏我的PHP代碼,
echo json_encode(array("package_name"=>$package_name,"package_type"=>$package_type,"discount"=>$discount,"amnt_discount"=>$amnt_discount,"tax"=>$tax,"amnt_tax"=>$amnt_tax,"count"=>$count,"job_post"=>$job_post,"package_price"=>$package_price,"valitity"=>$valitity));
你不是以json格式發送數據 – ManiMuthuPandi
它純粹在過去工作,但現在它返回錯誤 –
是否有任何問題,我的代碼 –