瞄准
删除选定ID的所有数据
问题
数据未被删除
错误
开发人员控制台中未提示任何错误
工作说明
根据我对代码功能的理解,检索用户ID(delete_userID
),然后使用此行将其传递到$delete_userID
。但我不明白为什么删除查询不起作用。我在网上看到了一些关于这个的问题,其中一些是因为数据库代码的丢失,而这个数据库代码是mysqli_real_escape_string($conn,$_POST['delete_userID']);
的,但这对我来说似乎不是问题。另外,我将其设置为一旦单击输入按钮,它将发送一个“delete”调用,并检查mysql_select_db("data2017", $conn);
语句是否已单击该按钮并且已收到该调用。除了检索用户ID之外,要删除的用户ID还通过使用JavaScript函数传入,如下所示。
删除代码
<div class="modal-body">
<p>Do you want to delete?</p>
</div>
<div class="form-group">
<label for="delete_userID" class="control-label">User ID:</label>
<input type="text" class="form-control" id="delete_userID" name="delete_userID" readonly/>
</div>
<div class="modal-footer">
<input type="submit" id="delete" name="delete" class="btn btn-block bt-login" value="Delete" />
<?php
if(isset($_POST['delete'])){
$delete_userID=mysqli_real_escape_string($conn,$_POST['delete_userID']);
mysql_select_db("data2017", $conn);
$sql = "DELETE FROM pha_user WHERE id ='".$delete_userID."'";
$Deletequery=mysqli_query($conn,$sql);
if($Deletequery){
echo "<script>alert('Delete Successful.');</script>";
}
else{
echo "<script>alert('Delete Failed.');</script>";
}
}
?>
</div>
更新
delete_userID是通过一个JavaScript函数传入的,如下所示。
function bindList() {
var $table = $('#eventsTable');
$table.bootstrapTable({
url: 'list-phaUser.php',
search: true,
pagination: true,
columns: [{
field: 'userID',
title: 'User ID',
sortable: true,
},{
field: 'operate',
title: 'Action',
align: 'center',
sortable: true,
events: operateEvents,
formatter: operateFormatter
},],
});
$(".fixed-table-toolbar").append("<div id='table-title' class='columns columns-left btn-group pull-left'>User List</div>");
$("#divFooter").remove();
$("<div class='row' id='divFooter' style='margin-left:-4%;margin-right:-4%;margin-bottom:-2%;'></div>").insertAfter("#divtb1");
$("#divFooter").load('footer.php');
}
window.operateEvents = {
'click .icon_edit': function (e, value, row, index) {
$("#edit_userID").val(row.userID);
$('#edit_model').modal('show');
},
'click .icon_delete': function (e, value, row, index) {
$("#delete_userID").val(row.userID);
$('#delete_model').modal('show');
}
};
最佳答案
你没有做表格。
<div class="modal-body">
<p>Do you want to delete?</p>
</div>
<form method="POST" action="">
<div class="form-group">
<label for="delete_userID" class="control-label">User ID:</label>
<input type="text" class="form-control" id="delete_userID" name="delete_userID" readonly/>
</div>
<div class="modal-footer">
<input type="submit" id="delete" name="delete" class="btn btn-block bt-login" value="Delete" />
</div>
</form>
<?php
if (isset($_POST['delete'])) {
$delete_userID = mysqli_real_escape_string($conn,$_POST['delete_userID']);
mysql_select_db("data2017", $conn);
$sql = "DELETE FROM pha_user WHERE id ='".$delete_userID."'";
$Deletequery = mysqli_query($conn,$sql);
if ($Deletequery) {
echo "<script>alert('Delete Successful.');</script>";
} else {
echo "<script>alert('Delete Failed.');</script>";
}
}
?>
关于php - PHP-SQL-DELETE查询不起作用,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/47989964/