我创建了一个包含发布数据链接的表,然后如果链接被点击,它将在新窗口中以我们想要的大小打开。
这是代码:
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['PONumber'] . "</td>";
echo "<td>" . $row['POdate'] . "</td>";
echo "<td>" . $row['customername'] . "</td>";
echo "<td>" . $row['description'] . "</td>";
echo "<td>" . $row['poqty'] . "</td>";
echo "<td>" . $row['TotalQtySpb'] . "</td>";
echo "<td>" . $row['OTSPO'] . "</td>";?>
<td><?php echo('<a href="index.php?action&ponumber='.$row['PONumber'].'" target="_blank">'."Klik Disini".'</a>');?></td><?php
echo "</tr>";
}
我试图通过在href中添加target =“_ blank”来修改open在新窗口中但是它不起作用(在新窗口中打开而不是在新窗口中打开)..
试过像这样的javascript但是不明白如何在表格中应用它。
<button onclick="myFunction()">Try it</button>
<script>
function myFunction() {
window.open("http://www.w3schools.com","","width=900,height=300,top=100,left=100" );
}
</script>
答案 0 :(得分:0)
Something like this应该让你去(未经测试,只是为了例如。):
<强> PHP:强>
$result = do_your_mysqli_query_here;
$out = '';
while($row = mysqli_fetch_assoc($result))
{
$out .= "<tr>";
$out .= "<td>" . $row['PONumber'] . "</td>";
$out .= "<td>" . $row['POdate'] . "</td>";
$out .= "<td>" . $row['customername'] . "</td>";
$out .= "<td>" . $row['description'] . "</td>";
$out .= "<td>" . $row['poqty'] . "</td>";
$out .= "<td>" . $row['TotalQtySpb'] . "</td>";
$out .= "<td>" . $row['OTSPO'] . "</td>";
$out .= '<td><a class="porow" href="#" data-ponumber="'.$row['PONumber'].'" target="_blank">Klik Disini</a></td>';
$out .= "</tr>";
}
echo $out;
<强>的javascript / jQuery的:强>
$(document).on('click', 'a.porow', function(){
var po = $(this).data('ponumber'); //gets value of data-ponumber for this a tag
var str = 'http://index.php?action=&ponumber=' + po;
window.location.href = str;
});
答案 1 :(得分:0)
可以使用jQuery
这样做$('td a').click(function(e){
// prevent browser following link
e.preventDefault();
//open in new window
window.open(this.href,"","width=900,height=300,top=100,left=100" );
});
向元素添加类并使用该类更新选择器