关于删除表格中的多条记录,我有一点问题。我使用复选框来删除它们但它不起作用。我不知道它的确切代码是什么。
这是我的PHP代码
<?php
echo "<form action='#'>
<fieldset>
<input type='text' name='search' value='' id='searchtalents' placeholder='Search Keywords..' size='40'/>
</fieldset>
</form>";
echo "<form name='tmsform' method='POST' action=''>";
$sql = "SELECT * FROM talentinfo WHERE 1 LIMIT 10";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ({$sql}) from DB: " . mysql_error();
exit;
}
if (mysql_num_rows($result) == 0) {
echo "No rows found";
exit;
}
echo"<div id='talentinfo'><table id='mr_database' name='myform' style='border:1px solid #fff;' cellspacing=0 cellpading='2' class='pretty'>
<thead>
<tr>
<th></th>
<th></th>
<th></th>
<th>Mr Tracking Number</th>
<th>First Name</th>
<th>Middle Name</th>
<th>Last Name</th>
<th>Address</th>
<th>Contact Number</th>
<th>School</th>
<th>Course</th>
<th>Year Graduated</th>
<th>Position Considered</th>
<th>Referred Location</th>
<th>Unit</th>
</tr>
</thead>";
$counter = 40000;
while ($row = mysql_fetch_assoc($result)) {
$filled = (($counter % 2) == 1) ? "style='background-color:#BCD9E1;'" : "" ;
$id = $row['talents_id'];
echo "<tbody><tr {$filled} class='tmsdel'>";
echo "<td><a href ='#' rel='#edit_talents{$counter}'><center><img src='img/edit.gif' width='25' height='21' title='Edit'></center></a></td>";
echo "<td><a href ='#' id=".$row['talents_id'].'&idelete=talents'." class='delete'><center><img src='img/delete.png' width='25' height='21' title='Delete'></center></a></td>";
echo "<td><input type='checkbox' name='checkbox[]' id='check".$row['talents_id']."' value='".$row['talents_id'].'&idelete=talents'."'/></td>";
echo "<td><a href='#' rel='#tracing_number{$counter}' style='text-decoration:none; font-weight:bold; color:#444;'>" . $row ['mr_tracking_number'] . "</a></td>";
echo "<td>" . $row['firstname'] . "</td>";
echo "<td>" . $row['middlename'] . "</td>";
echo "<td>" . $row['lastname'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>" . $row['contact_number'] . "</td>";
echo "<td>" . $row['school'] . "</td>";
echo "<td>" . $row['course'] . "</td>";
echo "<td>" . $row['year_graduated'] . "</td>";
echo "<td>" . $row['position_considered'] . "</td>";
echo "<td>" . $row['referred_location'] . "</td>";
echo "<td>" . $row['unit'] . "</td>";
echo "</tr></tbody>";
?>
这是我的Javascript
$(function(){
$(document).ready(function(){
});
$("#delete-all").click(function(){
$("#mr_database").remove();
$.ajax({
url: "modules/delete-all.php",
type: "get",
async: false,
data: "check"
});
});
});
答案 0 :(得分:1)
请按照以下步骤操作:
按钮上的调用功能单击以删除
var allVals = '';
$("#delete-all").click(function(){
$('.checkboxclass :checked').each(function() {
allVals = allVals + $(this).val() + ',';
});
}
然后您需要在allVals
中传递ajax
个变量并发布到.php文件
赞:数据:$ .ajax
'ids=' + allVals
最后,您可以在php文件中获取此变量,并对其执行删除过程。
喜欢:$ids = explode(',', $_REQUEST['ids');
并在mysql查询中使用id
希望这会对你有所帮助。
答案 1 :(得分:0)
相当简单的回答因为我在这里看到的是一个脚本来查看表并删除浏览器中的html而不是服务器上的数据。
您需要编写此处提到的modules / delete-all.php脚本: url:“modules / delete-all.php”
应包含“delete * FROM talentinfo”SQL查询。
然而,这将无缝删除所有表格内容。所以你可能想要提示:
$("#delete-all").click(function(){
Check = confirm("Do you really want to delete ALL DATA ? (Hint: there is no undo)");
if (Check == true) {
$("#mr_database").remove();
$.ajax({
url: "modules/delete-all.php",
type: "get",
async: false,
data: "check"
});
}
});
delete_all.php应该删除所有,所以只需执行:
<?php
$sql = "delete * FROM talentinfo ";
$result = mysql_query($sql);
if (!$result) {
echo "Could not successfully run query ({$sql}) from DB: " . mysql_error();
exit;
} else {
print "ok";
}
?>