我已经从数据库中获取数据,每件事都可以正常工作,但是问题是当我向test.php提交ajax请求时,我得到了每个按钮相同的值 我非常熟悉Ajax和Java,所以请帮助我,我混淆了如何分别获取每个按钮的价值并提交给test.php文件
/
这是我的test.php文件
<tbody>
<?php
$letter = mysqli_query($con,"SELECT * FROM letters order by id DESC");
if (mysqli_num_rows($letter) > 0) {
while ($rows_letter=mysqli_fetch_array($letter)) {
$id = $rows_letter['id'];
$subject = $rows_letter['subject'];
$status = $rows_letter['status'];
?>
<tr>
<th class="text-center" scope="row">1</th>
<td class="text-center"><?php echo $subject ;?></td>
<td class="text-center">
<?php
if ($status == 1) {
echo '<mark style="background-color: #5cb85c; color:white;"> Successfully Sent </mark>';
} else {
echo '<mark style="background-color:#f0ad4e; color:white;"> Not Sent Yet </mark>';
}
?>
</td>
<td>
<button type="button" class="btn btn-info btn-sm btn-block">
<span class="fa fa-pencil-square-o"></span> Edit</button>
</td>
<td>
<button type="button" class="btn btn-danger btn-sm btn-block">
<span class="fa fa-trash-o"></span> Move To Trash</button>
</td>
<td>
<button type="button" onclick="startsend();" id="id" value="<?php echo $id;?>"class="btn btn-success btn-sm btn-block">
<span class="fa fa-paper-plane-o"></span> Send To All</button>
</td>
</tr>
<?php
}
}
?>
</tbody>
<script type='text/javascript'>
//AJAX function
function startsend() {
var id = $('#id').val();
$.ajax({
type: "POST",
url: "test.php",
data:{ id: id
},
success: function(msg){
alert( "Button Id is " + msg );
}
});
}
</script>
答案 0 :(得分:1)
尝试此操作,将ID作为param
传递给ajax
HTML:
<td><button type="button" onclick="startsend(<?php echo $id;?>);"
id="id" value="<?php echo $id;?>"class="btn btn-success btn-sm btn-block">
<span class="fa fa-paper-plane-o"></span> Send To All</button></td>
Ajax:
function startsend(id) {
$.ajax({
type: "POST",
url: "test.php",
data:{ id: id },
success: function(msg){
alert( "Button Id is " + msg );
}
});
}