PHP中回调JQuery $.post中的函数,传递1个参数我就知道,但传递多个参数该如何写?
如以下例子:$.post("/home/getcode",function(cl7,cl8){var cl9=cl7.SecCode;var cla=cl7.KCode;var clb=camelcrypto(cl9,cla);$.post("/home/vote",{id:cl6,selOpts:cl4,questName:cl5,chkCode:$("#chkcode").val(),secCode:clb,num1:Math.random()},function(clc,cld){$("#btn_vote").show();var cle=clc.RCode;$("#vote_tips").text("");if(cle==-1000){$("#chkCodeMsg").text(clc.Msg);$("#chkCodeDialog").dialog("open");$("#chkCodeDialog a").click()}else{alert(clc.Msg);if(clc.Success){document.location=document.location}}})})};cl2.preventDefault()});$("#btn_votecode").click(function(){$("#btn_vote").click()})});function showVoteBar(){$(".vcount").show();$(".votebar :hidden").each(function(cl0){var cl1=$(this).val();$(this).parent().animate({"width":cl1+"px"},1000)})}
PHP中该如何返回调用
如以下例子:$.post("/home/getcode",function(cl7,cl8){var cl9=cl7.SecCode;var cla=cl7.KCode;var clb=camelcrypto(cl9,cla);$.post("/home/vote",{id:cl6,selOpts:cl4,questName:cl5,chkCode:$("#chkcode").val(),secCode:clb,num1:Math.random()},function(clc,cld){$("#btn_vote").show();var cle=clc.RCode;$("#vote_tips").text("");if(cle==-1000){$("#chkCodeMsg").text(clc.Msg);$("#chkCodeDialog").dialog("open");$("#chkCodeDialog a").click()}else{alert(clc.Msg);if(clc.Success){document.location=document.location}}})})};cl2.preventDefault()});$("#btn_votecode").click(function(){$("#btn_vote").click()})});function showVoteBar(){$(".vcount").show();$(".votebar :hidden").each(function(cl0){var cl1=$(this).val();$(this).parent().animate({"width":cl1+"px"},1000)})}
PHP中该如何返回调用
把参数放到表单中去,直接 用jquery 的param = $("#myform").serialize();
令 dataType = json
则 jq 就会按 json 处理数据
php 端只需 echo json_encode(...);在你无格式的代码中,怎么也没有找到需要多参数的地方
改为
function(d) {
c17 = d.c17;
c18 = d.c18;php:
echo json_encode(array('c17' => 123, 'c18' => 456));jq 不可能预知你要怎么用它,所以才用了统一格式来处理
就所谓:以不变应万变