Question

我尝试使用jQuery调用函数，该函数使用GET发送参数。此函数返回JSON结果。

这是我的代码：

 $(document).ready(function() {
     function getUrlVars() {
         var vars = [], hash;
         var hashes = window.location.href.slice(window.location.href.indexOf('?') + 1).split('&');
         for (var i = 0; i < hashes.length; i++) {
           hash = hashes[i].split('=');
           vars.push(hash[0]);
           vars[hash[0]] = hash[1];
         }
         return vars;
     }

     var nom_restaurant = decodeURI(getUrlVars()["nom_restaurant"]);
      //alert(nom_restaurant);
      $.ajax({
          type: 'GET',
          url: 'http://192.168.1.15/fou/affiche.php',
          data:nom_restaurant ,

          success: function (data) {
                //Do stuff with the JSON data
                $.getJSON(url,data,function(result){
                  console.log(result);
                  $.each(result, function(i, field){

                     var codeBarre=field.codeBarre;
                     var designation=field.designation;
                     var dateDebut=field.dateDebut;
                     var dateFin=field.dateFin;
                     var prix=field.prix;
                     var TVA=field.TVA;
                     var photo=field.photo;

                    $("#promo").append(" "+"<a href='form.html?codeBarre="+codeBarre+"&designation="+designation+"&dateDebut="+dateDebut+"&dateFin="+dateFin+"&prix="+prix+"&TVA="+TVA+"&photo="+photo+"'><h2>"+designation+"</h2>"+" "+dateDebut+" "+dateFin+" "+'<img src="http://192.168.1.15/fou/phonegapserver/phonegapserver/uploads/'+photo+'" width=50px height=50px />'+"<br/></a>");
                });
              });
          }
        });
  });

我的方法调用：

   $("#list").append("<a href='liste.html?nom_restaurant="+TABLE_NAME+"'>"+TABLE_NAME+"</a><br/>");

我的功能列出了一个表格内容。我在php上的功能：

function AffichePromo($nom_restaurant){

$db1=new PDO('mysql:host=localhost;dbname=testf','root','');


$db1->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
$sql = 'SELECT codeBarre,designation,dateDebut,dateFin,prix,TVA,photo FROM '.$nom_restaurant.'';
      foreach($db1->query($sql)as $data)
    {

       $texte[]=$data;

    }
         return json_encode($texte);

    }

我使用肥皂网服务，餐馆名称来自。

Answer 1

这是我的解决方案：

$(document).ready(function() {
    function getUrlVars() {
        var vars = [], hash;
        var hashes = window.location.href.slice(window.location.href.indexOf('?') + 1).split('&');
        for (var i = 0; i < hashes.length; i++) {
            hash = hashes[i].split('=');
            vars.push(hash[0]);
            vars[hash[0]] = hash[1];
        }
        return vars;
    }

    var nom_restaurant = decodeURI(getUrlVars()["nom_restaurant"]);

    $.getJSON('http://192.168.1.15/fou/affiche.php', {nom_restaurant : nom_restaurant }, function(result) {
        $.each(result, function(i, field) {
            var codeBarre=field.codeBarre;
            var designation=field.designation;
            var dateDebut=field.dateDebut;
            var dateFin=field.dateFin;
            var prix=field.prix;
            var TVA=field.TVA;
            var photo=field.photo;

            $("#promo").append(" "+"<a href='form.html codeBarre="+codeBarre+"&designation="+designation+"&dateDebut="+dateDebut+"&dateFin="+dateFin+"&prix="+prix+"&TVA="+TVA+"&photo="+photo+"'><h2>"+designation+"</h2>"+" "+dateDebut+" "+dateFin+" "+'<img src="http://192.168.1.15/fou/phonegapserver/phonegapserver/uploads/'+photo+'" width=50px height=50px />'+"<br/></a>");

        });
    });

});

我如何使用pass参数从jQuery.getJSON获取callBack

1 个答案: