通過彈出窗口在url中跳轉空格和特殊字符

Question

我的彈出窗口工作正常，直到變量中有一些空格或特殊字符（如＃等） 例如，如果我的引用變量類似於'12345'但是當涉及空格和特殊字符時，它不起作用。 我正在使用php結合javascript彈出窗口來查看項目的每個參考。

<script type="text/javascript"> 
function PopupCenter(pageURL, title,w,h) 
{ 
var left = (screen.width/2)-(w/2); 
           var top = (screen.height/2)-(h/2); 
           var targetWin = window.open (pageURL, title, 'toolbar=no, location=no, directories=no, status=no, menubar=no, scrollbars=no, copyhistory=no, width='+w+', height='+h+', top='+top+', left='+left); 
          } 
          </script> 


    <?php 
    $sql = "SELECT * from purchase_tb ORDER BY ponumber ASC "; 
    $result = $conn->query($sql); 
    if ($result->num_rows > 0) { 
    echo " 
    <div class='table-responsive'> 
    <table class='table table-bordered table-hover table-condensed '> 
    <tr class='success'><th>PO.Number</th><th> Model </th><th>Date Created</th> 
    </tr>"; 
    $current_cat = null; 
    while($row = $result->fetch_assoc()) { 
    if ($row["ponumber"] != $current_cat) { 
    $current_cat = $row["ponumber"]; 
    $current_cat2 = $row["datecreated"]; 
    echo " 
    <tr><td>"; 
    echo" 
    {$current_cat}</td> 
    <td><a href='#' onclick=PopupCenter('viewpurchase.php?Rid=".$row['ponumber']."','myPop1',1000,800)>View</a></td><td>{$current_cat2}</td>"; 
?> 

Answer 1

URL必須是... URL編碼。

像空格這樣的字符必須替換爲有效字符才能形成有效的URL。

例如：

viewpurchase.php ID = 2 334

成爲

viewpurchase.php％3Fid％3D2％20334

嘗試PHP函數urlencode在代碼：

<a href='#' onclick=PopupCenter('viewpurchase.php?Rid=". 

urlencode($row['ponumber']). 

"','myPop1',1000,800)>View</a></td><td>{$current_cat2}</td>";