2013-07-06 28 views
1

抱歉,我對這個問題沒有具體的標題,window.open印刷,並通過調用數據庫字段

我想打印與window.open方法一種形式,這是一些一段腳本:

<?php 
include "connect.php"; 

$test = mysql_query("select * from dataorder order by idorder desc limit 1 "); 


if (!test) 
die ("Error... " .mysql_error()); 

while ($row = mysql_fetch_array ($test)) 
{ 
echo "ID number : $row[idorder] <br/><br/>"; 
echo "Name  : $row[name] <br/><br/>"; 
echo "<div>"; 
echo "<input type=button name=back value='back'onClick='parent.location='reserve.php''>"; 
echo '<input type="submit" 
      class="btn btn-success" 
      name="print" value="print" 
     onClick="window.open("print.php?idorder="'.$row['idorder'].'","scrollwindow","top=200,left=350,width=680px,height=500")" 
     style="text-decoration:none">'; 
echo "</div>"; 
} 
?> 

「打印」按鈕可以出現,但點擊時沒有響應。

回答

1

我覺得你有防止onClick無法正常工作的多個報價的問題(加上空間問題) -

echo "<input type=button name=back value='back'onClick='parent.location='reserve.php''>"; 
       ^ ^^^   ^         ^
echo '<input type="submit" 
     class="btn btn-success" 
     name="print" value="print" 
    onClick="window.open("print.php?idorder="'.$row['idorder'].'","scrollwindow","top=200,left=350,width=680px,height=500")" 
         ^    ^    ^^   ^^          ^
    style="text-decoration:none">'; 

嘗試這樣的事情 -

echo '<input type="button" name="back" value="back" onClick="parent.location=\'reserve.php\'">'; 
echo '<input type="submit" class="btn btn-success" name="print" value="print" 
     onClick="window.open(\'print.php?idorder='.$row['idorder'].'\',\'scrollwindow\',\'top=200,left=350,width=680px,height=500\')" 
    style="text-decoration:none">'; 
+0

感謝,它的工作原理:) – andrian

+0

沒有問題。樂意效勞。 – Sean