通過PHP從MySql傳遞URL

Question

我正在努力解決如何傳遞存儲在MySQL數據庫中的URL並將其附加到圖像。 劇本我到目前爲止是這樣的：

<?php 
$con = mysqli_connect("**********","*********","******","********"); 

// Check connection 
if (mysqli_connect_errno()) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} 
$selectedOption = $_POST["mySelect"]; 

$result = mysqli_query($con, 
    sprintf("SELECT * FROM `SouthYorkshire` WHERE `EstProv` = '%s'", 
     preg_replace("/[^A-Za-z ]/", '', $selectedOption) 
    ) 
); // pattern based on your html select options 

echo "<div id=\"Results\">"; 

while($row = mysqli_fetch_array($result)) { 
    echo "<div class=\"ClubName\">"; 
    echo $row['EstName']; 
    echo "</div><br>"; 
    echo "<div class=\"Location\">"; 
    echo $row['EstAddress2']; 
    echo "</div>"; 
    echo "<br>"; 
    echo "<div id=\"website\"><img src=\"photos/visit-website-button.png\" width=\"75\" height=\"25\" /></div>"; 
} 
echo date("Y") . " " ."Search is Powered by PHP."; 
echo "</div>"; 

mysqli_close($con); 

我試圖改變該位爲： echo "<div id=\"website\"><img src=\"photos/visit-website-button.png\" width=\"75\" height=\"25\" /></div>"

我試着用上述同樣的用$row['EstWebsite']

，但我沒有任何成功，任何建議都會很棒。

非常感謝

Answer 1

哪裏並不重要的URL從何而來，到底它只是一個數據的字符串，你只是要正在建設一些HTML，所以

echo '<div id="website"><a href="' . $url . '"><img blah blah blah></a>"; 
         ^^^^^^^^^^^^^^^^^^^^^^^-add this   ^^^^--add this 

Answer 2

您可以使用以下方法：

echo "<div id=\"website\"><img src=\"" . $row['ImageColumnName'] ."\" width=\"75\" height=\"25\" /></div>";