[英]sent dynamic table to mysql
我正在尝试将动态表发送到mysql数据库,但是遇到了困难。 我尝试使用for进行发送,但距离还不太远。 目前,我只是显示它,但我真的想将其发送到mysql。 我希望所有数据与其余数据(如test case name
, test case number
等)合而为一。 但是动态表必须放在其余数据所在的同一行中。
在下面,您将看到我做了什么,也许您可以看到我的意图是...
<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 $user_username ?>" readonly/>
<label>System:</label>
<input class="form-control" type="text" name="test_case_system"/>
<label>URL:</label>
<input class="form-control" type="text" name="test_case_url"/>
</div>
<div class="col-md-12" style="border: 1px solid #28415b; padding-bottom: 12px; padding-top: 12px;margin-top: 20px; margin-bottom: 15px">
<p>
<INPUT class="btn btn-primary ladda-button" type="button" value="Add row" onclick="addRow('dataTable')" />
<INPUT class="btn btn-primary ladda-button" type="button" value="Delete row" onclick="deleteRow('dataTable')" />
</p>
<div class="clear"></div>
<p>'All fields below are compatible to use markdowns for editing' </p>
<table class="table table-hover">
<thead>
<tr>
<th style="width: 15px">Chk</th>
<th style="width: 335px">Action:</th>
<th style="width: 326px;">Expected System Response:</th>
<th style="width: 151px;">Pass/ Fail</th>
<th>Comment</th>
</tr>
</thead>
</table>
<table id="dataTable" class="table table-hover">
<tbody>
<tr>
<td style="width:20px;"><INPUT type="checkbox" name="chk[]" id="chk"/></td>
<td><INPUT class="form-control" type="text" name="step[]" autocomplete="on" placeholder="Action" required/></td>
<td><INPUT class="form-control" type="text" name="url[]" autocomplete="on" placeholder="Expected Outcome" required/></td>
<td>
<select name="passfail[]" class="form-control" style="width:120px;">
<OPTION value="Pass">....</OPTION>
<OPTION value="Pass">Pass</OPTION>
<OPTION value="Fail">Fail</OPTION>
</select>
</td>
<td>
<TEXTAREA class="form-control" type="text" name="comment[]" rows="2" cols="15" placeholder="Comment" required></TEXTAREA>
</td>
</tr>
</tbody>
</table>
那就是我动态制作的桌子...
在我的PHP文件中,我显示动态表,如下所示:
<table class="table table-bordered">
<thead>
<tr>
<td>Step</td>
<td>process</td>
<td>Expected System Response</td>
<td>
<center>Pass/ Fail</center>
</td>
<td>Comment</td>
</tr>
</thead>
<?php
if (isset($_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>
现在如何将其插入mysql中???
因此,基本上,需要将其插入数据库中是在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);
因此,您必须打开sql连接,然后将数据插入到所需的表中。 我希望这足够清楚。 现在,如果调用了PHP文件,则会发出SQL查询。
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