这是从数据库 r_job 表中获取的大量列表。 当我单击“查看详细信息”按钮时，它必须通过 ajax 将作业值带到 Controller 页面，然后我将获取作业信息，作为响应，我必须发送与作业相关的表字段，并且必须在此页面上显示。

这是我的动态列表:

$Jobquery = $conn->query("SELECT * FROM r_job ");
while($JobResults = $Jobquery->fetch_assoc()){

<tr>
    <td id="hiringevent"><?php echo $JobResults['hiringevent']; ?></td>
    <td id="JobId"><?php echo $JobResults['id_job']; ?></td>
    <td><button id="ViewDetails" class="btn btn-primary text-center">View Details</button></td>
</tr>

这是我的 ajax 和 jquery 调用:

$("#ViewDetails").click(function() {
    $.ajax({
            url: "job-controller.php",
            method: "POST",
            data: {'action':'viewjob','JobId' : + $('#JobId').html()},
            dataType: "json",
            success: function (response) {
                $("#showMessage").html(response['message']);
            },
            error: function (request, status, error) {
                $("#showMessage").html("OOPS! Something Went Wrong Please Try After Sometime!");
            }
        }); 
        return false;
    });

最后是我的 Controller 页面:

if($_POST['action']=='viewjob'){ 
        $jobSearch= $conn->query("SELECT * From r_job WHERE id_job='".$_POST['JobId']."'") or die(mysql_error());
        $ViewJob=$jobSearch->fetch_assoc();
        $hiringevent        =   $ViewJob['hiringevent'];
        $jobname            =   $ViewJob['jobname'];
        $jobdescription     =   $ViewJob['jobdescription'];
        $cutoff             =   $ViewJob['cutoff'];
        $joblocation        =   $ViewJob['joblocation'];
        $interviewlocation  =   $ViewJob['interviewlocation'];
        $jobexperience      =   $ViewJob['jobexperience'];
            $response['message'] = "Show Job Information";
            $response['success'] = true;
            }else{
            $response['message'] = "OOPS! Something Went Wrong Please Try After Sometime!";
            $response['success'] = false;
        }
        echo json_encode($response);
        exit;
    }

My current issue is when i click view details only first view details button working remaining no response

最佳答案

请注意，每页应该有一个 id。 您需要做的就是，创建一个 js 函数并在单击按钮时调用它，并将 JobId 作为参数发送。当你在循环中时，它工作正常。

HTML

$Jobquery = $conn->query("SELECT * FROM r_job ");
while($JobResults = $Jobquery->fetch_assoc()){

<tr>
    <td id="hiringevent"><?php echo $JobResults['hiringevent']; ?></td>
    <td id="JobId"><?php echo $JobResults['id_job']; ?></td>
    <td><button class="btn btn-primary text-center" onClick="getDetails(<?php echo $JobResults['id_job']; ?>)">View Details</button></td>
</tr>
}

JS

function getDetails(jobID){
// ajax Call Here
console.log(jobID)// you can see JobID Here.
// now set the values to by using id
$("#ViewJobId").val(response['jobData']['jobId']);
/* val is used to set aswell as get the values */



  /*in case of td you have to use text insted of val*/
   $("#ViewJobId").text(response['jobData']['jobId']);

}

关于php - 我如何发送 JobId 并在 View 中 append 值。 Ajax ，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/39831821/