如何发送JobId并在视图上追加值.阿贾克斯

Question

这是从数据库r_job表中获取的lotal列表..当我单击查看详细信息按钮时,它必须通过ajax将作业值带到控制器页面,然后我将获取作业信息,并且在响应中我必须发送作业相关表feillds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;
    });

最后我的控制器页面:

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;
    }

我当前的问题是,当我单击查看详细信息时,只有第一个查看详细信息按钮仍然没有响

Answer 1

请注意，每个页面应该有一个 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。当你处于循环中时效果很好。

超文本标记语言

$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']);

}