Question

我想检查用户的专业知识并显示他们的网络图。我想使用LIKE的原因是因为在mysql数据库中，其工作场所还包含了专有技术属性。例如：Petronas的数据分析师。因此，我想检查他们的专业知识是否是数据分析师，并显示他们的网络图。

这是我尝试过的代码：

<?php
if (isset($_GET['id'])) {

    $id = $_GET['id'];
    $sql = "SELECT * FROM professional, job, location WHERE PROFESSIONAL_ID LIKE '%$id%' AND PROFESSIONAL_ID=JOB_ID AND JOB_ID = LOCATION_ID " ;
    $result = mysqli_query($conn, $sql);

    $row = mysqli_fetch_assoc($result);

    echo "
                <div class='card'>
                <img src='../images/img.png' style='width:100%'>
                <h2>".$row['PROFESSIONAL_NAME']."</h2>
                <p class='title'><i class='fa fa-briefcase' aria-hidden='true'></i> ".$row['JOB_NAME']."</p>
                <p class='marker'><i class='fa fa-map-marker' aria-hidden='true'></i> ".$row['LOCATION_NAME']."</p>
                <a href=".$row['PROFESSIONAL_URL']."><p><button>Contact</button></p></a>
              </div>

              <div class='network'> 
              <h2> Filter Network</h2>
              </div>

              <div class='filter'>
        <input type='radio' value='Expertise' unchecked name='radioBtn' onclick='checkexpertise(".$row['JOB_NAME'].")'> <label> Expertise</label><br>
        <input type='radio' value='Location' unchecked name='radioBtn' onclick='checklocation(".$row['LOCATION_NAME'].")'> <label> Location</label><br>
        <input type='radio' value='Workplace' unchecked name='radioBtn' onclick='checkworkplace()'> <label> Workplace </label><br>
        <input type='radio' value='Past Workplace' unchecked name='radioBtn' onclick='checkpast()'> <label>Past Workplace</label><br>

      </div>";


}

?>
<script>
      function CheckExpertise (Expertise) {
        if Expertise LIKE %Data Analyst% OR %data analyst%
        {
          window.location.replace("expertise.html");
          }
      }
      </script>

Answer 1

您可以使用includes，我建议将字符串转换为所有小写字母，然后再进行检查，因此您只需要进行一次检查即可。

function CheckExpertise (Expertise) {
    if (Expertise.toLowerCase.includes("data analyst") {
        window.location.replace("expertise.html");
    }
}

Answer 2

您可以为此使用正则表达式：

if (/.*data analyst.*/i.test(Expertise)) {
  window.location.replace("expertise.html");
}

但是请看一下mysqli准备好的语句和输出清理（例如使用htmlentities）。用户可以将任何内容发送为“ id”，包括您可能不想执行的精巧SQL查询；）

$id = $_GET['id'];
$query = mysqli_prepare($conn, "SELECT * FROM professional, job, location WHERE PROFESSIONAL_ID LIKE ? AND PROFESSIONAL_ID=JOB_ID AND JOB_ID = LOCATION_ID " ;
mysqli_bind_param($query, "s", "%{$id}%");
mysqli_stmt_execute($query);
$result = mysqli_stmt_get_result($query);

如何在javascript中的mysql中执行某些操作？

2 个答案: