Javascript 如何在ajax中获取数组值

我试图在Ajax中显示值，我在employee.php页面中获取数组值，但当我将值传递给getdata.js时，它没有显示结果

请检查以下结果以了解此：：>

 echo json_encode($data);   [{"0":"2","tax_id":"2","1":"GST","tax_type":"GST","2":"1","tax_comp":"1","3":"10","tax_Percent":"10"},{"0":"3","tax_id":"3","1":"CGST","tax_type":"CGST","2":"1","tax_comp":"1","3":"9","tax_Percent":"9"},{"0":"8","tax_id":"8","1":"new child","tax_type":"new child","2":"1","tax_comp":"1","3":"15","tax_Percent":"15"}] 

-------getEmployee.php------------------------

             if($_REQUEST['tax_id']) {
             $sql = "SELECT tax_id, tax_type, tax_comp, tax_Percent FROM ca_taxmaster WHERE tax_comp='".$_REQUEST['tax_id']."'";
              $resultset = mysql_query($sql) or die(mysql_error());
              $data = array();
              while( $rows = mysql_fetch_array($resultset) ) {
              $data[] = $rows;
             // $data[] = $rows['tax_id'] . " " . $rows['tax_type'] . " " . $rows['tax_comp'] . " " . $rows['tax_Percent'];
              }
              echo json_encode($data);
              } else {
              echo 0;
              }?>

  <select id="employee">
    <option value="" selected="selected">Please select</option>

    <?php
    $sql = "SELECT tax_id, tax_type, tax_comp, tax_Percent FROM ca_taxmaster where tax_comp = '0'";
    $resultset = mysql_query($sql) or die( mysqli_error());
    while( $rows = mysql_fetch_array($resultset)) {
    ?>
    <option value="<?php echo $rows["tax_id"]; ?>"><?php echo $rows["tax_type"]; ?></option>
    <?php } ?>
   </select>


  <div id="display">
  <div class="row" id="heading" style="display:none;"><h3><div class="col-sm-4"><strong>Employee Name</strong></div><div class="col-sm-4"><strong>Age</strong></div><div class="col-sm-4"><strong>Salary</strong></div></h3></div><br>      
  <div class="row" id="records">
    <div class="col-sm-4" id="tax_type"></div>
    <div class="col-sm-4" id="tax_comp"></div>
    <div class="col-sm-4" id="tax_Percent"></div>

  </div>     
  <div class="row" id="no_records"><div class="col-sm-4">Plese select employee name to view details</div></div>

-------------------getData.js------------------------------------ 路径：-resources/js/getData.js>

    $(document).ready(function(){
   // code to get all records from table via select box
   $("#employee").change(function() {
   var tax_id = $(this).find(":selected").val();
   var dataString = 'empid='+ tax_id;
   $.ajax({
        url: 'http://localhost/capms_v2_feb/ajax/getEmployee.php',
        dataType: "json",
        data: dataString,
        cache: false,
        success: function(employeeData) {
        //alert(data);
   if(employeeData) {

       $("#heading").show();
       $("#no_records").hide();
       $("#tax_type").text(employeeData.tax_type);
       $("#tax_comp").text(employeeData.tax_comp);
       $("#tax_Percent").text(employeeData.tax_Percent);
       $("#records").show();
    } else {
       $("#heading").hide();
       $("#records").hide();
       $("#no_records").show();
        }
      }
     });
    })
   });

-------------------看法------------------

             if($_REQUEST['tax_id']) {
             $sql = "SELECT tax_id, tax_type, tax_comp, tax_Percent FROM ca_taxmaster WHERE tax_comp='".$_REQUEST['tax_id']."'";
              $resultset = mysql_query($sql) or die(mysql_error());
              $data = array();
              while( $rows = mysql_fetch_array($resultset) ) {
              $data[] = $rows;
             // $data[] = $rows['tax_id'] . " " . $rows['tax_type'] . " " . $rows['tax_comp'] . " " . $rows['tax_Percent'];
              }
              echo json_encode($data);
              } else {
              echo 0;
              }?>

  <select id="employee">
    <option value="" selected="selected">Please select</option>

    <?php
    $sql = "SELECT tax_id, tax_type, tax_comp, tax_Percent FROM ca_taxmaster where tax_comp = '0'";
    $resultset = mysql_query($sql) or die( mysqli_error());
    while( $rows = mysql_fetch_array($resultset)) {
    ?>
    <option value="<?php echo $rows["tax_id"]; ?>"><?php echo $rows["tax_type"]; ?></option>
    <?php } ?>
   </select>


  <div id="display">
  <div class="row" id="heading" style="display:none;"><h3><div class="col-sm-4"><strong>Employee Name</strong></div><div class="col-sm-4"><strong>Age</strong></div><div class="col-sm-4"><strong>Salary</strong></div></h3></div><br>      
  <div class="row" id="records">
    <div class="col-sm-4" id="tax_type"></div>
    <div class="col-sm-4" id="tax_comp"></div>
    <div class="col-sm-4" id="tax_Percent"></div>

  </div>     
  <div class="row" id="no_records"><div class="col-sm-4">Plese select employee name to view details</div></div>

如果有人知道，请帮助我，提前谢谢

请按以下连结：

---

您可以这样做：

var employeeData=[{0:2，税号：2,1:GST，税号：GST，2:1，税号：1,3:10，税号：10}，{0:3，税号：3,1:CGST，税号：CGST，2:1，税号：1,3:9，税号：9，{0:8，税号：8,1：新生儿，税号：新生儿，税号：2:1，税号：1,3:15]]； employeeData.forEachfunctionitem{ var数据=； 数据+=+项目.税种+； 数据+=+项目税+公司税； 数据+=+项目税百分比+； 数据+=； $'.appendData'.appendData； }; 类型 营地 百分率

---

您应该尝试添加headerContent类型：application/json；在对php文件进行json_编码之前

请尝试更改以下行：

 if($_REQUEST['tax_id']) to $_REQUEST['empid']
    $sql = "SELECT tax_id, tax_type, tax_comp, tax_Percent FROM ca_taxmaster WHERE tax_comp='".$_REQUEST['tax_id']."'"; 
to 
$sql = "SELECT tax_id, tax_type, tax_comp, tax_Percent FROM ca_taxmaster WHERE tax_comp='".$_REQUEST['empid']."'";

OR try this



 Change var dataString = 'empid='+ tax_id; to  var dataString = 'tax_id ='+ tax_id;

---

您应该尝试添加headerContent类型：application/json；在你的php文件上。你会受到sql注入攻击，顺便说一句：@Jennifer你能运行console.logemployeeData；关于你的成功ajax，看看它显示了什么？@Jennifer我发现你的情况是检查$u请求['tax\u id']，这是不正确的。将其更改为$_REQUEST['empid']谢谢cjatstackoverflow它现在正在工作$_REQUEST['empid']已更正，但我以这种方式请求值数组var employee=[employeeData]；employeeData.forEachfunctionitem{var data=；data+=+item.tax_type+；data+=+item.tax_comp+；data+=+item.tax_Percent+；data+=；$'.appendData'.appendData；}；现在很好，谢谢你time@Jennifer如果这能解决您的问题，请将此设置为答案：