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我寫了一個php代碼,當我點擊提交按鈕時,組合框中的一些項目將被刪除。現在我想要確認,並且我在下面的代碼中寫下了這個不工作 PHP代碼:JQuery確認對話框提交點擊PHP
$DeleteButton=$_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的部分:
<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>
問題是這樣的代碼是行不通的。如果我沒有添加jquery部分,代碼完全正常工作,但添加jquery部分後,當我點擊提交按鈕時,jquery對話框出現,但點擊yes按鈕後,表單將被提交而不刪除選定的項目。
請你詳細說明更多。我只是想刪除按鈕點擊項目。我怎樣才能使用這個隱藏的領域來實現呢? – Amir
謝謝。那個答案救了我^ _ ^ – Amir