php - 发送动态表到mysql

Question

我正在尝试将动态表发送到 mysql 数据库，但遇到困难。我尝试使用 for 来发送它，但我没有走得太远。目前我只是显示它，但我真的想将它发送到 mysql。我希望所有数据与其余数据(例如测试用例名称、测试用例编号等)位于一行中。但动态表必须放置在其余数据所在的同一行中。

下面你会看到我做了什么，也许你能明白我的意图......

<div class="col-md-6">
  <label>Test Case Number:</label>

<?php
  $sql = "SELECT test_case_number FROM qa_testing_application ORDER BY id desc LIMIT 1";
  $result = mysqli_query($database, $sql) or trigger_error("SQL", E_USER_ERROR);

  while ($row = mysqli_fetch_assoc($result)) { 
?>
  <input class="form-control" style="display: none" type="text" name="test_case_number" readonly="readonly" value="<?php echo $row['test_case_number']+3?>">
<?php } ?>
  <label>Company Name:</label>
  <input class="form-control" type="text"  name="test_case_company_name"/>
  <label>Tester:</label>
  <input class="form-control" style="display: none" type="text"  name="user_username" value="<?php echo $user_username ?>" readonly/>
  <label>System:</label>
  <input class="form-control" type="text"  name="test_case_system"/>
  <label>URL:</label>
  <input class="form-control" type="text"  name="test_case_url"/>
</div>

<div class="col-md-12" style="border: 1px solid #28415b; padding-bottom: 12px; padding-top: 12px;margin-top: 20px; margin-bottom: 15px">
  <p>
      <INPUT class="btn btn-primary ladda-button" type="button" value="Add row" onclick="addRow('dataTable')" />
      <INPUT class="btn btn-primary ladda-button" type="button" value="Delete row" onclick="deleteRow('dataTable')" />
  </p>
  <div class="clear"></div>
  <p>'All fields below are compatible to use markdowns for editing' </p> 



<table class="table table-hover">
    <thead>
    <tr>
        <th style="width: 15px">Chk</th>
        <th style="width: 335px">Action:</th>
        <th style="width: 326px;">Expected System Response:</th>
        <th style="width: 151px;">Pass/ Fail</th>
        <th>Comment</th>
    </tr>
    </thead>
</table>
<table id="dataTable" class="table table-hover">
    <tbody>
    <tr>
        <td style="width:20px;"><INPUT type="checkbox" name="chk[]" id="chk"/></td>
        <td><INPUT class="form-control" type="text" name="step[]"  autocomplete="on" placeholder="Action" required/></td>
        <td><INPUT class="form-control" type="text" name="url[]"  autocomplete="on" placeholder="Expected Outcome" required/></td>
        <td>
            <select name="passfail[]" class="form-control" style="width:120px;">
                <OPTION value="Pass">....</OPTION>
                <OPTION value="Pass">Pass</OPTION>
                <OPTION value="Fail">Fail</OPTION>
            </select>
        </td>
        <td>
            <TEXTAREA class="form-control" type="text" name="comment[]" rows="2" cols="15" placeholder="Comment" required></TEXTAREA>
        </td>
    </tr>
    </tbody>
</table>

这是我动态制作的表格...

在我的 PHP 文件中，我显示动态表如下:

<table class="table table-bordered">
    <thead>
          <tr>
          <td>Step</td>
          <td>process</td>
          <td>Expected System Response</td>
          <td>
              <center>Pass/ Fail</center>
          </td>
          <td>Comment</td>
          </tr>
    </thead>
<?php
    if (isset($_POST)) {
    $step = $_REQUEST['step'];
    $url = $_REQUEST['url'];
    $pass_fail = $_REQUEST['passfail'];
    $comment = $_REQUEST['comment'];
    $countPass = 0;
    $countFail = 0;
    foreach ($step as $key => $row) {
?>
    <tbody>
    <tr>
        <td><?php echo $key + 1; ?></td>
        <td><?php echo $step[$key]; ?></td>
        <td><?php echo $url[$key]; ?></td>
        <td style="color:<?php if ($pass_fail[$key] == 'Fail') {
            echo 'color: red';
        } else {
            echo 'limegreen';
        } ?>"><b>
                <center><?php echo $pass_fail[$key]; ?></center>
            </b></td>
        <td><?php echo $comment[$key]; ?>
        </td>
    </tr>
    </tbody>
</table>

现在我如何将它插入mysql???

Answer 1

所以基本上，将其插入数据库所需要做的就是在 PHP 中执行 sql 查询。

让我向您展示一个代码示例:

//establish db connection
$con=mysqli_connect("dbhost","username","dbpassword","dbname");

$sql = "INSERT INTO tablename (name_of_row1, 
                               name_of_row2, 
                               name_of_row3) 
                       VALUES ('".$value1."',
                               '".$value2."',
                               '".$value3."')";

mysqli_query($con, $sql);
mysqli_close($con);

所以你必须打开一个sql连接，然后将你的数据插入到你想要的表中。我希望这已经足够清楚了。现在，如果 PHP 文件被调用，就会进行 SQL 查询。