$(document).on('mouseenter', '.grid-img-hover', function() {
var container = $(this);
var jobId = container.parent().find('.title-wrap-hidden').text();
$.ajax({
url: 'db_client_job_name_lookup.php',
type: 'POST',
dataType: 'json',
data:{
jobId: jobId
},
success: function(data) {
//console.log(data.client_name[0][0][0]['name']);
container.find('.title-wrap').html('<p class="job-name">'+ data.portfolio_item_name +
'</p><p class="client-name">'+ data.client_name[0][0][0]['name'] +'</p>');
// container.find('.title-wrap').html('<p class="client-name">'+ data.result.client_name +'</p>');
},
error: function(jqXHR, textStatus, errorThrown) {
// error
alert(errorThrown);
}
});
});
您好,我有这行代码,我想循环遍历data.client_name[0][0][0]['name']
并在同一<p class="client-name"></p>
标记中输出其值。索引会在data.client
的中间改变,如此(data.client_name[0][$index][0]['name'])
。我怎么能这样做?感谢。
答案 0 :(得分:0)
这是你的意思吗?
for (i = 0; i < data.client_name[0].length; i++) {
container.find('.title-wrap').html('<p class="job-name">'+ data.portfolio_item_name +
'</p><p class="client-name">'+ data.client_name[0][i][0]['name'] +'</p>');
}
答案 1 :(得分:0)
$(document).on('mouseenter', '.grid-img-hover', function() {
var container = $(this);
var jobId = container.parent().find('.title-wrap-hidden').text();
$.ajax({
url: 'db_client_job_name_lookup.php',
type: 'POST',
dataType: 'json',
data: {
jobId: jobId
},
success: function(data) {
var client_names_text;
var clients_array = [];
function make_clients(){
for (var i = 0; i < data.client_name[0].length; i++) {
clients_array.push(data.client_name[0][i][0]['name']);
}
client_names_text = clients_array.join(',');
return client_names_text;
}
//console.log(data.client_name[0][0][0]['name']);
container.find('.title-wrap').html('<p class="job-name">'+ data.portfolio_item_name +
'</p><p class="client-name">'+ make_clients() +'</p>');
// container.find('.title-wrap').html('<p class="client-name">'+ data.result.client_name +'</p>');
},
error: function(jqXHR, textStatus, errorThrown) {
// error
alert(errorThrown);
}
});
});
好的,我通过创建一个功能来完成繁重的工作,这里有工作代码块。