我有一个表单，用户可以在其中选择烛台的引用。
选择完成后，与烛台相关的事件列表会出现在 Bootstrap 表中。
该表由 Mysql 数据库中的项目完成。
一切正常。现在我想在另一个名为“测试”的列中添加一个按钮。但是我没有按钮..我放了一张我看到的截图..
我的代码是这样的:
表格文件
<?php
require_once 'login.php';
$sql= "SELECT ptlum FROM ptlum where ptlum LIKE '%AZ%'";
$result = mysql_query($sql) or die("Requete pas comprise");
//while($data = mysql_fetch_array($result))
// {
//echo "<option>".$data[ptlum]."</option>";
// }
?>
<html>
<head>
<meta charset="utf-8">
<link href="examples.css" rel="stylesheet">
<link href="http://minikomi.github.io/Bootstrap-Form-Builder/assets/css/lib/bootstrap.min.css" rel="stylesheet">
<link href="http://minikomi.github.io/Bootstrap-Form-Builder/assets/css/lib/bootstrap-responsive.min.css" rel="stylesheet">
<link href="http://minikomi.github.io/Bootstrap-Form-Builder/assets/css/custom.css" rel="stylesheet">
<link href="bootstrap.table.css" rel="stylesheet">
<script>
function showUser(str) {
if (str == "") {
document.getElementById("txtHint").innerHTML = "";
return;
} else {
if (window.XMLHttpRequest) {
// code for IE7+, Firefox, Chrome, Opera, Safari
xmlhttp = new XMLHttpRequest();
} else {
// code for IE6, IE5
xmlhttp = new ActiveXObject("Microsoft.XMLHTTP");
}
xmlhttp.onreadystatechange = function() {
if (xmlhttp.readyState == 4 && xmlhttp.status == 200) {
document.getElementById("txtHint").innerHTML = xmlhttp.responseText;
}
}
xmlhttp.open("GET","getuser.php?q="+str,true);
xmlhttp.send();
}
}
</script>
</head>
<body>
<form class="form-horizontal" >
<legend>Rechercher une panne en cours ou archivée</legend>
<div class="control-group">
<label class="control-label" for="selectbasic-0">Sélectionner un point lumineux</label>
<div class="controls">
<select name="users" onchange="showUser(this.value)">
<?php
while ($row = mysql_fetch_array($result))
{
echo "<option>".$row[ptlum]."</option>";
}
?>
</select>
</div>
</div>
</form>
<script src= "jquery.js"></script>
<script src= "bootstrap.min.js"></script>
<script src= "bootstrap.table.js"></script>
<br>
<div id="txtHint"><b></b></div>
</body>
</html>
调用数据库的 PHP 文件
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<link href="https://cdn.datatables.net/1.10.9/css/jquery.dataTables.min.css" rel="stylesheet">
</head>
<body>
<script src= "jquery.js"></script>
<script src= "https://cdn.datatables.net/1.10.9/js/jquery.dataTables.min.js"></script>
<script src= "https://cdn.datatables.net/1.10.9/js/dataTables.bootstrap.min.js"></script>
<?php
$q = $_GET['q'];
//echo $q;
$con = mysqli_connect('localhost','root','root','sdeer');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM depannages WHERE ptlum = '".$q."'";
$result = mysqli_query($con,$sql);
echo "<table id=\"example\" class=\"table table-striped table-bordered\" cellspacing=\"0\" width=\"100%\">
<tr>
<th>Pt Lum</th>
<th>Matériel</th>
<th>Prestation</th>
<th>Date</th>
<th>Nature</th>
<th>Test</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tbody>";
echo "<tr>";
echo "<td>" . $row['ptlum'] . "</td>";
echo "<td>" . $row['materiel'] . "</td>";
echo "<td>" . $row['presta'] . "</td>";
echo "<td>" . $row['date'] . "</td>";
echo "<td>" . $row['nature'] . "</td>";
echo "</tr>";
echo "</tbody>";
}
echo "</table>";
mysqli_close($con);
?>
<script>
$(document).ready(function() {
var table = $('#example').DataTable( {
"columnDefs": [ {
"targets": -1,
"data": null,
"defaultContent": "<button>Click!</button>"
} ]
} );
</script>
</body>
</html>
如果我想要一个按钮，如何制作？当用户单击该按钮时，将出现一个新窗口。
谢谢!
最佳答案
您可以使 anchor 标记看起来像按钮。只需添加 role="button"。
<a href="Give-Some-Page-URL-Here.php" class="btn btn-default" role="button" target="blank">New Window Button</a>
编辑代码:
<!DOCTYPE html>
<html>
<head>
<meta charset="utf-8">
<link href="https://cdn.datatables.net/1.10.9/css/jquery.dataTables.min.css" rel="stylesheet">
<script src= "jquery.js"></script>
<script src= "https://cdn.datatables.net/1.10.9/js/jquery.dataTables.min.js"></script>
<script src= "https://cdn.datatables.net/1.10.9/js/dataTables.bootstrap.min.js"></script>
</head>
<body>
<?php
$q = $_GET['q'];
$con = mysqli_connect('localhost','root','root','sdeer');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"ajax_demo");
$sql="SELECT * FROM depannages WHERE ptlum = '$q'";
$result = mysqli_query($con,$sql);
?>
<table id="example" class="table table-striped table-bordered" cellspacing="0" width="100%">
<thead>
<tr>
<th>Pt Lum</th>
<th>Matériel</th>
<th>Prestation</th>
<th>Date</th>
<th>Nature</th>
<th>Test</th>
</tr>
</thead>
<tbody>
<?
while($row = mysqli_fetch_array($result))
{?>
<tr>
<td><?echo $row['ptlum'];?></td>
<td><?echo $row['materiel'];?></td>
<td><?echo $row['presta'];?></td>
<td><?echo row['date'];?></td>
<td><?echo $row['nature'];?></td>
<td>
<a href="Give-Some-Page-URL-Here.php" class="btn btn-default" role="button" target="blank">New Window Button</a>
</td>
</tr>
<?}?>
<tbody>
</table>
<?mysqli_close($con);?>
<script>
$(document).ready(function() {
var table = $('#example').DataTable( {
"columnDefs": [ {
"targets": -1,
"data": null,
"defaultContent": "<button>Click!</button>"
} ]
} );
</script>
</body>
</html>
关于php - 包含 MySQL 内容的 Bootstrap 数据表，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/33263518/