2016-11-07 28 views
1

我是一名學生(很新的PHP)導入的HTML元素的背景顏色試圖動態地改變每個「魚」的背景顏色中數據庫,但在此之前,我需要嘗試靜態更改背景,但是,我甚至無法做到這一點。我試圖用引號,引號和轉義字符來引用無效。非更新內嵌CSS時,試圖改變從MySQL數據庫

PHP/HTML

<html> 
<head> 
    <title>Dinosaurs</title> 
</head> 
<body> 
</body> 
    <?php 
    include 'connection.php'; 
    $query = "SELECT Name FROM `fish` WHERE FishTypeID = 4"; 
    if (!$result = $mysqli->query($query)){ 
    echo "Sorry, the website is experiencing problems."; 
    echo "Error: Your query failed to execute and here is why: \n"; 
    echo "Errno: " . $mysqli->errno . "\n"; 
    echo "Error: " . $mysqli->error . "\n"; 
    exit; 
    } 
    while ($row = mysqli_fetch_row($result)) { 
    print("<div background-color = \"blue\"> 
      <h4>"); 
    foreach ($row as $key => $value) { 
     print($value . " 
       <svg height = 64 width = 80> 
        <polygon points='1 31, 6 42, 1 47,26 63,60 50,80 64, 72 39, 80 16,60 28,26 17' /> 
        <circle cx='20' cy='32' r='5' stroke='white' stroke-width='3' fill='black' /> 
       </svg>"); 
    } 
    print("</h4></div>"); 
    } 
    ?> 
    <h1>dinosaurs</h1> 
</html> 
+0

你是對的,當你說你首先需要做沒有數據庫,但是這個代碼與數據庫組件 – e4c5

+0

@ e4c5我有一個String嘗試過,但沒有奏效。 –

+0

是你爲什麼通過添加數據庫相關的東西使它更復雜? – e4c5

回答

0

感謝所有幫助!事實證明,我爲我的背景顏色分配使用了「=」而不是「:」。