JQuery确认对话框提交点击PHP

时间:2016-11-26 11:28:41

标签: php jquery jquery-ui form-submit jquery-ui-dialog

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。

1 个答案:

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;