我写了一个PHP代码,当我点击提交按钮时,组合框中的某些项目将被删除。现在我想要确认,我在下面编写的代码无法正常工作。 php代码:
$DeleteButton=$_REQUEST['DeleteButton'];
if ($DeleteButton=="delete") :
if ($DeleteComboBox=="PickOne") :
$DeleteButton = "" ;
else :
$query = "DELETE FROM `items` WHERE `id` = $DeleteComboBox LIMIT 1";
$result = mysql_query($query)
or die("Database deletion failed");
$DeleteButton = "" ;
endif ;
endif ;
echo "<BR><BR><FORM NAME=\"EditFORM\" ACTION=\"./index.php\" METHOD=POST>\n";
$sql_select = "SELECT * FROM items WHERE id>0 order by name" ;
$sql_result = mysql_query($sql_select)
or die ("Couldn't execute SQL query on db table.") ;
echo "<SELECT ID=\"DeleteComboBox\" NAME=\"DeleteComboBox\">";
echo "<OPTION VALUE=\"PickOne\" SELECTED>select item</OPTION>";
while ($row = mysql_fetch_array($sql_result)) {
echo "<OPTION VALUE=\"$row[0]\">" . $row[2] . " " . $row[1] . "</OPTION>";
}
echo "</SELECT>";
echo "<BR><BR><INPUT TYPE=SUBMIT NAME=\"DeleteButton\" VALUE=\"delete\" ID=\"DeleteButton\">\n" ;
echo "</FORM>\n";
JQuery部分:
<script type="text/javascript">
$(document).ready(function() {
$("#dialog").dialog({
autoOpen: false,
modal: true
});
});
$("#DeleteButton").click(function(e) {
e.preventDefault();
currentForm = $(this).closest('form');
$("#dialog").dialog({
dialogClass: "no-close",
buttons : {
"yes" : function() {
currentForm.submit();
},
"no" : function() {
$(this).dialog("close");
}
}
});
$("#dialog").dialog("open");
});
</script>
问题是此代码无效。如果我没有添加jquery部分代码完全正常工作但是在我单击提交按钮后添加jquery部分时会出现jquery对话框,但是在单击yes按钮后,将提交表单而不删除所选项目。
答案 0 :(得分:1)
提交按钮的值仅在单击时提交,但您捕获此事件并执行e.preventDefault()
。在它之后,currentForm.submit()
不再记得点击了哪个按钮。
您可以动态地向表单添加隐藏的输入:
currentForm.append('<input type="hidden" name="action" value="delete" />');
currentForm.submit()
而不是检查$_REQUEST['DeleteButton']
,您可以在PHP中检查此隐藏输入的值:
$action = $_REQUEST['action'];
if ($action == 'delete'):
// ... delete the item
endif;