为什么此构造函数不起作用？ （在Ajax成功中）

Question

I'm trying to extract my json data and put into a variable, which is available from everywhere. But I've got an error message, it says : foods is undefined (the row of the alert in the end)

  var foods;
          function search() {
            $.ajax({
              url: "foodsrequest.php",
              type: "GET",
              dataType: "json",
              async: false,
              data: {"inputData": JSON.stringify(filterdata)},
              success: function(data){

                foods = foodConstructor(data[0]); ///yes, it is an array of objects and it has all the parameters needed
                function foodConstructor(dataIn){
                  this.id = dataIn.id;
                  this.name = dataIn.name;
                  this.price = dataIn.price;
                  this.species = dataIn.species;
                  this.type = dataIn.type;
                  this.manufacturer = dataIn.manufacturer;
                  this.weight = dataIn.weight;
                  this.age = dataIn.age;
                  this.partner = dataIn.partner;
                }
              }
            });
          }

          alert(foods.name);

Answer 1

You forgot the new keyword

Try:

            foods = new foodConstructor ( data[ 0 ]); ///yes, it is an array of objects and it has all the parameters needed function foodConstructor ( dataIn ){ this . id = dataIn . id ; this . name = dataIn . name ; this . price = dataIn . price ; this . species = dataIn . species ; this . type = dataIn . type ; this . manufacturer = dataIn . manufacturer ; this . weight = dataIn . weight ; this . age = dataIn . age ; this . partner = dataIn . partner ; } } }); } 

      alert ( foods . name ); 

Answer 2

Just try invoking your constructor with new keyword. It will work.

foods = new foodConstructor(data[0]);

Answer 3

Howard Fring what you are going to want to do is move the alert into a function and call it from the call back function that is called when the ajax request succeeds. food is not populated until the request is finished that is why it is undefined. For example:

var foods;
      function search() {
        $.ajax({
          url: "foodsrequest.php",
          type: "GET",
          dataType: "json",
          async: false,
          data: {"inputData": JSON.stringify(filterdata)},
          success: function(data){

            foods = new foodConstructor(data[0]); ///yes, it is an array of objects and it has all the parameters needed
            function foodConstructor(dataIn){
              this.id = dataIn.id;
              this.name = dataIn.name;
              this.price = dataIn.price;
              this.species = dataIn.species;
              this.type = dataIn.type;
              this.manufacturer = dataIn.manufacturer;
              this.weight = dataIn.weight;
              this.age = dataIn.age;
              this.partner = dataIn.partner;
            }
            foodAlert();
          }
        });
      }

      function foodAlert(){
        alert(foods.name);
      }

By calling foodAlert in the success call back after food is populated will open an alert with the value of food.name shown.