將超鏈接添加到回顯結果的PHP搜索腳本中

Question

我有以下搜索腳本，我需要將公司部分的結果鏈接到每個公司網站（存儲在我的數據庫的WEBSITE列中）。誰能幫忙？

if(isset($_POST['completedsearch'])) { 
    $term = $_POST['query']; 
    $mysql = mysql_connect("ldb504.securepod.com","martindb","green11"); 
    mysql_select_db("hcsd"); 
    $qu = mysql_query("SELECT * FROM Sheet1 WHERE COMPANY LIKE '%".mysql_real_escape_string($term)."%' OR LOCATION LIKE '%".mysql_real_escape_string($term)."%' OR KEYWORDS LIKE '%".mysql_real_escape_string($term)."%' OR PRODUCTSSERVICES LIKE '%".mysql_real_escape_string($term)."%' "); //selects the row that contains ANYTHING like the submitted string 
    echo "Searching for '$term'"; 
    echo " 
     <table><tr style=\"border-bottom:1px dotted #305896;\"><th style=\"padding:0px 10px;text-align:left;border-bottom:1px dotted #305896;\">Company</th> 
     <th style=\"padding:0px 10px;text-align:left;border-bottom:1px dotted #305896;\">Location</th> 
     <th style=\"padding:0px 10px;text-align:left;border-bottom:1px dotted #305896;\">Products/Services</th></tr>"; 
    while($row = mysql_fetch_array($qu)) { 
     echo"<tr style=\"border-bottom:1px dotted #305896;\"><td style=\"padding:0px 10px;text-align:left;border-bottom:1px dotted #305896;\">"; 
     echo"<a href="'.urlencode($row['WEBSITE']).'">'.htmlspecialchars($row['COMPANY']).'</a>"; 
     echo "</td>"; 
     echo "<td style=\"padding:0px 10px;text-align:left;border-bottom:1px dotted #305896;\">"; 
     echo htmlspecialchars($row['LOCATION']); 
     echo "</td>"; 
     echo "<td style=\"padding:0px 10px;text-align:left;border-bottom:1px dotted #305896;\">"; 
     echo htmlspecialchars($row['PRODUCTSSERVICES']); 
     echo "</td></tr>"; 
    } 
    echo "</table>"; 
} 

Answer 1

你在你的PHP混合單引號和雙引號的錯誤：

echo"<a href="'.urlencode($row['WEBSITE']).'">'.htmlspecialchars($row['COMPANY']).'</a>"; 

應該是這樣的：

echo '<a href="'.urlencode($row['WEBSITE']).'">'.htmlspecialchars($row['COMPANY']).'</a>';