PHP MySQL的where子句不讀取變量

Question

我在這里開始生氣

mysql_query即使我嘗試將其重命名也無法識別我的變量$d1

這是代碼。

HTML：

<form action ="manageVessel.php" method ="POST">
  <select onchange ="this.form.submit();" class ="form-control" name ="ViewPositionCertificates">
      <option>Choose a Position </option>                                                       
      <?php
      $ViewPCertificates = mysql_query("SELECT * FROM table_cmsjob") or die("error" . mysql_error());
      while ($rwViewPCertificates = mysql_fetch_array($ViewPCertificates)) {
          ?>
          <option value =" <?php echo $rwViewPCertificates['jobName']; ?> "> <?php echo $rwViewPCertificates['jobName']; ?></option>  
      <?php } ?>

       </select>   
</form>

的PHP：

   <?php if (isset($_POST['ViewPositionCertificates'])) { ?>
      <table class = "table table-bordered">
          <tr class ="bg-primary">
              <td> List of Certificates </td>
          </tr>
          <?php
          $d1 = $_POST['ViewPositionCertificates'];
          echo $_POST['ViewPositionCertificates'];
          $ViewCertificatesFP = mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName =  '$d1' ") or die("error" . mysql_error());

          while ($rwViewCertificatesFP = mysql_fetch_array($ViewCertificatesFP)) {

              echo "<tr>";
              echo "<td>" . $rwViewCertificatesFP['Certificate'] . "</td>";
              echo "</tr>";
          }
          ?>

      </table>




  <?php } ?>

例如，當我使用字符串時，MYSQL WHERE子句工作正常

    mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName =  'MASTER' ") or die("error" . mysql_error());

但是當我使用變量將$_POST['ViewPositionCertificates']分配給變量MYSQL WHERE子句時，它沒有幫助嗎？

Answer 1

<option value ="<?php echo $rwViewPCertificates['jobName']; ?>"> <?php echo $rwViewPCertificates['jobName']; ?></option>   // remove xtra spaces from here......

從value屬性中刪除空格

Answer 2

像這樣嘗試

$ViewCertificatesFP = mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName =  '".mysql_real_escape_string(trim($d1))."' ") or die("error" . mysql_error());

要么

$ViewCertificatesFP = mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName =  '".addslashes(trim($d1))."' ") or die("error" . mysql_error());

Answer 3

<?php if (isset($_POST['ViewPositionCertificates'])) { ?>
    <table class = "table table-bordered">
        <tr class ="bg-primary">
            <td> List of Certificates </td>
        </tr>
        <?php
        $d1 = $_POST['ViewPositionCertificates'];
        echo $_POST['ViewPositionCertificates'];
        $ViewCertificatesFP = mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName =  '".$d1."' ") or die("error" . mysql_error());

        while ($rwViewCertificatesFP = mysql_fetch_array($ViewCertificatesFP)) {

            echo "<tr>";
            echo "<td>" . $rwViewCertificatesFP['Certificate'] . "</td>";
            echo "</tr>";
        }
        ?>
    </table>
<?php } ?>

選擇查詢語法中的更改使用單引號。

Answer 4

試試這個：

mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName =  ".$_POST['ViewPositionCertificates'])

Answer 5

您不能在單引號內編寫變量，如果您將其編寫，則PHP會將其視為字符串。 因此您的查詢將是
$ViewCertificatesFP = mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName = $d1 ") or die("error" . mysql_error());

如需更多幫助，請閱讀PHP中的變量插值

Answer 6

嘗試更改您的查詢方法可能會有所幫助...

$ViewCertificatesFP = mysql_query("SELECT * FROM table_cmsjobassigning WHERE jobName =  '".$d1."' ") or die("error" . mysql_error());