Question

我面临一些问题：

注意：未定义的变量：第44行的C：...中的searchResult

警告：mysqli_fetch_array（）要求参数1为mysqli_result，在第44行的C：...中给出为null

我不知道如何解决它，我们将非常感谢任何帮助。感谢

generatereport.php

<?php
if(isset($_POST['search']))
{
    $courseid = $_POST['courseid'];
    $query = "SELECT * FROM course_registration WHERE r_cid ='$courseid'";
    $searchResult = filterTable($query);
}

function filterTable($query)
{
    include("include/config.php");
    $filter_Result = mysqli_query($link, $query);
    return $filter_Result;
}
?>

<fieldset style="font-family: 'Exo', sans-serif;">
<legend>Generate Report</legend>
<form name="GenForm" method="post" action="generatereport.php" onSubmit="return InputCheck(this)">
<table>
    <tr>
        <td>Course ID:</td>
        <td><input id="courseid" name="courseid" type="text" class="input" pattern="([A-Z]{3,4})-[A-Z0-9]{3,5}" 
        title="Please enter a valid course id">
        <span>(eg. DPT-3054)</span></td>
    </tr>
</table>

    <div>
        <input type="submit" name="search" value="  Search By Course ID  ">
    </div>

<table width="70%" cellpadding="5" cellspacing="5">
<tr>
   <th><strong>Course ID</strong></th>
   <th><strong>Full Name</strong></th>
   <th><strong>Student ID</strong></th>
   <th><strong>Program</strong></th>
   <th><strong>Intake</strong></th>
</tr>

<?php while ($row = mysqli_fetch_array($searchResult)) :?> **//line 44
<tr>
   <td><?php echo $row['r_cid']; ?></td>
   <td><?php echo $row['r_name']; ?></td>
   <td><?php echo $row['r_sid']; ?></td>
   <td><?php echo $row['r_program']; ?></td>
   <td><?php echo $row['r_fintake']; ?></td>
</tr>
<?php endwhile; ?>
</table>

</form>
</fieldset>

<script language=JavaScript>

function InputCheck(GenForm)
{
  if (GenForm.courseid.value == "")
  {
    alert("The field Course ID cannot be left blank!");
    GenForm.courseid.focus();
    return (false);
  }
}

</script>

的config.php

<?php

    $link= mysqli_connect("localhost","root","","course_registration_system");

?>

Answer 1

在顶部使用

if(isset($_POST['search']))
{
    $courseid = $_POST['courseid'];
    $query = "SELECT * FROM course_registration WHERE r_cid ='$courseid'";
    $searchResult = filterTable($query);
}

因此，如果设置了$ _POST [＆＃39; search＆＃39;]，则只会设置$ searchResult。稍后，如果还设置了此值，则还应仅处理此值。因此，当if（$ _ POST [＆＃39; search＆＃39;]）

时，表的生成应该只在一个块内。

Answer 2

  use this code. i am sure it will be work
<?php
if(isset($_POST['search']))
{
    $courseid = $_POST['courseid'];
    $query = "SELECT * FROM course_registration WHERE r_cid ='$courseid'";
    $searchResult = filterTable($query);
}

function filterTable($query)
{
    include("include/config.php");
    $display_list=array();
    $filter_Result = mysqli_query($link, $query);
   while ($row = mysqli_fetch_array($filter_Result)) 
   {
     $display_list[]=$row;
   }
    return $filter_Result;
}
?>

<fieldset style="font-family: 'Exo', sans-serif;">
<legend>Generate Report</legend>
<form name="GenForm" method="post" action="generatereport.php" onSubmit="return InputCheck(this)">
<table>
    <tr>
        <td>Course ID:</td>
        <td><input id="courseid" name="courseid" type="text" class="input" pattern="([A-Z]{3,4})-[A-Z0-9]{3,5}" 
        title="Please enter a valid course id">
        <span>(eg. DPT-3054)</span></td>
    </tr>
</table>

    <div>
        <input type="submit" name="search" value="  Search By Course ID  ">
    </div>

<table width="70%" cellpadding="5" cellspacing="5">
<tr>
   <th><strong>Course ID</strong></th>
   <th><strong>Full Name</strong></th>
   <th><strong>Student ID</strong></th>
   <th><strong>Program</strong></th>
   <th><strong>Intake</strong></th>
</tr>

<?php 
if(!empty($searchResult)){
foreach($searchResult as $row) { ?>
<tr>
   <td><?php echo $row['r_cid']; ?></td>
   <td><?php echo $row['r_name']; ?></td>
   <td><?php echo $row['r_sid']; ?></td>
   <td><?php echo $row['r_program']; ?></td>
   <td><?php echo $row['r_fintake']; ?></td>
</tr>
<?php }} ?>
</table>

</form>
</fieldset>

<script language=JavaScript>

function InputCheck(GenForm)
{
  if (GenForm.courseid.value == "")
  {
    alert("The field Course ID cannot be left blank!");
    GenForm.courseid.focus();
    return (false);
  }
}

</script>

警告：mysqli_fetch_array（）期望参数1为mysqli_result，在第44行的C：\ xampp \ htdocs \ ...中给出null

2 个答案: