Question

我是一个PHP新手。我正在创建一个工作网站，我的搜索功能告诉我什么时候没有结果但是如果有，它会显示我在数据库中输入的所有工作。请协助，我已经尝试了一切。

这是我的代码：

 <?php


    if(isset($_POST['submit']))
   {

      $search = $_POST['keyword'];

      $query = "SELECT * FROM jobs WHERE job_tags LIKE '%$search%'";
      $search_query = mysqli_query($connection, $query);

      if(!$search_query) {
        die ("query failed" . mysqli_error($connection));
      }

      $count = mysqli_num_rows($search_query);
    if($count == 0){
        echo "<h3> NO RESULT</h3>";
    }else{

    $query = "SELECT * FROM jobs"; 
    $job_display = mysqli_query($connection, $query);

    while($row = mysqli_fetch_assoc($job_display)){
        $job_title = $row['job_title'];
        $employer = $row['employer'];
        $job_date = $row['job_date'];
        $job_logo = $row['job_logo'];
        $job_desc = $row['job_desc'];

?>
<div class="row">
    <div>



                <div class="media img-responsive">
                    <div class="media-left media-middle">

                        <a href="#">
                        <img class="media-object" src="images/<?php echo $job_logo; ?>" class="img-responsive" alt="Absa Insurance Logo">
                        </a>
                     </div>
                        <div class="media-body">
                        <h4 class="media-heading"><span class="job-tittle"><?php echo "{$job_title}";?> </span>(<i class="glyphicon glyphicon-map-marker"> </i>Gauteng, <span class="type blue"> Short-Term Insurance</span>)</h4>
                        <P>
                        <?php echo $job_desc;?>  
                        ...<a href="details.html"> <i class="glyphicon glyphicon-plus"> </i> Read More</a></P>

                    </div>

                        <div class=" media-right media-middle job-location">

                       <p> <?php echo $job_date;?> </p>

                        </div>
                </div>
    </div>
</div>

                        <?php } 

                                    }
                            }

?>

这是表格

 <form class=" form-inline" action="search.php" method="post">
                            <div class="form-group">
                                <input type="text" name="keyword" class="form-control" placeholder="Job Key Word">
                            </div>
                            </form>

如果您需要更多信息，请与我们联系。

Answer 1

您需要删除此行或只是在此处过滤

$query = "SELECT * FROM jobs";

Answer 2

在第一个查询中您只查找选定的记录

$query = "SELECT * FROM jobs WHERE job_tags LIKE '%$search%'";
                            $search_query = mysqli_query($connection, $query);

如果找到了您再次搜索的内容

$query = "SELECT * FROM jobs";

您不会在此查询中放置WHERE。

我的PHP搜索显示数据库上的所有条目，即使只有一个是正确的

2 个答案: