Question

我在加载页面时调用web服务，即其下面的代码包含在addmachine.php的顶部

<?php
session_start();
$url='webservice url';
    $ch=curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($useridofuser));
    curl_setopt($ch, CURLOPT_HTTPHEADER, array("Content-Type: application/json"));
    $response=  curl_exec($ch);
    echo('\n'."Server response : \n \n".$response);
    curl_close($ch);
    //parsing the json response from server
    $jsonde="$response";
    $_SESSION['json']=$jsonde;
  $org = array();
  $loc = array();
  $bui = array();
  $items = json_decode($response);
  foreach( $items as $each ){
 $loc[]=$each->location[0]->name;
 $bui[]=$each->location[0]->building[0];
 $org[]=$each->name;
 }
 ?>

Web服务的响应是json如下

[{"location":[{"building":["Default Building"],"name":"Default Location"}],"name":"Default Organization"},{"location":[{"building":["test_loc1_building1","test_loc1_building2"],"name":"test location1"},{"building":["test_loc2_building2"],"name":"test location2"}],"name":"test Organization"}]

我已经解码了json，并在组织，位置和建筑物的下拉菜单中作为选项移动，然后在从下拉列表中选择组织，构建和位置后，需要使用ajax调用Web服务

   <form name="theForm" method="post" action="addmachine.php" enctype="multipart/form-data" onSubmit="return validate();">
          <label for="orgname">Organisation Name</label>
                    <select style="width: 305px;text-align:left ;"  name="category_id" id="category_id">
                    <option value="">Select</option>
                    <?php foreach($org as $key=>$val){?>
                    <option value="<?php echo $val; ?>"><?php echo $val;?></option>
 <?php
 }
 ?>                      </select>

                    <p>
        <label name="location">Location</label>

                     <select style="width: 305px;" name="category_id1" id="category_id1">
                     <option value="">Select</option>
                     <?php foreach($loc as $key=>$val){?>
                     <option value="<?php echo $val; ?>"><?php echo $val; ?></option>
 <?php
 }
?>
                     </select>
                    </p>
                    <p>
        <label for="building">Building</label>

                    <select style="width: 305px" name="category_id2" id="category_id2">
                    <option value="">Select</option>
                    <?php foreach($bui as $key=>$val){?>
                    <option value="<?php echo $val; ?>"><?php echo $val; ?></option>
 <?php
 }
 ?>
                    </select>
                    </p>
                    <label for="entr/exi">Entrance/Exit</label>
                    <select style="width: 305px;text-align:left ;"  name="building" id="category_id1" onkeypress="return ajax();">
                    <option value="">Select</option>
                    <?php foreach($org as $key=>$val){?>
                    <option value="<?php echo $val; ?>"><?php echo $val;?></option>
    <?php
    }
    ?>                      </select>
                    <p>
                    <label for="type">Type</label>
                    <input type="text" name="type" placeholder="enter your work station"/>

          <label for="name">Name</label>
          <input type="text" name="workstnname" placeholder="enter your work station" onkeypress="return onKeyPressBlockNumbers(event);">
          <label for="description">Description</label>
          <textarea name="description" style="height:150px;width:300px;"></textarea>
          <label for="machinetype">Machine Type</label>
                    <select style="width: 305px;text-align:left;"  name="machinetype">
                    <option value="">Select</option>
                    <option value="kiosk">kiosk</option>
                    <option value="workstation">workstation</option>

                  </select>
                    <p>
                    <input type="submit" name="submit" value="Submit">
                    </p>

        </form>

因为我已经调用了函数ajax（）;关于字段入口/出口的按键。我已经在这里进行了ajax调用，我希望执行另一个名为ajaxweb.php的php页面我的ajax代码如下所示在页面的脚本标签中addmachine.php

            $.ajax({
                type: "GET",        //GET or POST or PUT or DELETE verb
                url:ajaxweb.php,        // Location of the service
                data: "",       //Data sent to server
                contentType: "json",        // content type sent to server
                dataType: "json",   //Expected data format from server
                processdata: true,  //True or False
                success: function (json) {//On Successful service call
                    var result = json.name;
                    $("#dvAjax").html(result);
                },
                error: ServiceFailed    // When Service call fails
            });

            return false;
        });

ajaxweb.php，其中选定的组织，位置和建筑物被提供给Web服务以在同一页面上返回，但它不起作用我也想知道如何将该响应发送到addmachine.php页面。

 <?php
 session_start();
 $org=$_POST['category_id'];
 $loc=$_POST['category_id1'];
 $bui=$_POST['category_id2'];
 $addmachine=array(
'org_name'=>$org,
'loc_name'=>$loc,
'building_name'=>$bui
 );
 echo json_encode($addmachine);
 $url='webservice url';
    $data="$addmachine";
     echo("Input to Server : ".$data."\n");
    $ch=curl_init();
    curl_setopt($ch, CURLOPT_URL, $url);
    curl_setopt($ch, CURLOPT_RETURNTRANSFER, true);
    curl_setopt($ch, CURLOPT_POST, true);
    curl_setopt($ch, CURLOPT_POSTFIELDS, json_encode($addmachine));
    curl_setopt($ch, CURLOPT_HTTPHEADER, array("Content-Type: application/json"));
    $response=  curl_exec($ch);
    echo('\n'."Server response : \n \n".$response);
    curl_close($ch);
    ?>

希望有人能提供帮助谢谢

Ajax调用没有返回任何东西

0 个答案: