我不知道JS的位置可能会改变我的html页面的结果,直到今天.我希望图像src在两个不同的URL"点击"之间切换.为什么这个第一个代码的工作方式与我想要的一样,但第二个代码却没有？第二个代码的源html为var not_a_bad_word生成一个空字符串.
第一个代码:
<!DOCTYPE HTML>
<html>
<head>
<meta charset="UTF-8">
<title>'Murica!'</title>
</head>
<body>
<?php
$dbhost = 'databasePlace';
$dbname = 'mine';
$dbuser = 'me';
$dbpass = '*****';
$link = mysqli_connect($dbhost, $dbuser, $dbpass, $dbname);
mysqli_select_db($link, $dbname);
$name = $_GET["fname"];
$query = sprintf(
"SELECT image_url, Type
FROM Pokemon c
WHERE c.name = '%s'",
mysqli_real_escape_string($link, $name));
$result = mysqli_fetch_assoc(mysqli_query($link, $query));
echo '<img id="pokemon_card" onclick="changeImage()" height="450"
width="330" src="' . $result['image_url'] . '"/>';
mysqli_close($link);
?>
<script>
function changeImage() {
element = document.getElementById('pokemon_card');
var not_a_bad_word = "<?php echo …我刚刚开始使用html,php和mysql.我已经使用php成功登录了我的数据库,并形成了一个查询.我现在想更进一步,展示一张图片而不仅仅是字符串或数字.
变量'result'将返回一个字符串,该字符串包含我想要在此网页上显示的图像的url.我该怎么做？
<html>
<head>
<title>My First Question</title>
</head>
<body>
<?php
$dbhost = 'someURL.com';
$dbname = 'user';
$dbuser = 'user';
$dbpass = 'password';
$mysql_handle = mysql_connect($dbhost, $dbuser, $dbpass)
or die("Error Connecting To Database Server");
mysql_select_db($dbname, $mysql_handle)
or die("Error selecting database: $dbname");
echo 'Successfully connected to database!';
$first = 'bobbie';
$query = sprintf("SELECT image FROM Player
p WHERE name='%s'", mysql_real_escape_string($first));
$result = mysql_query($query);
mysql_close($mysql_handle);
?>
</body>
</html>
我查找与宠物小精灵相关的图像并用php显示它.然后我希望能够通过点击它"翻转卡片".我第一次点击下来,但第二次点击将卡翻转回来是行不通的.我认为它是JS中我的php变量的语法:
<!DOCTYPE HTML>
<html>
<head>
<meta charset="UTF-8">
<title>
'Murica!
</title>
<script>
function changeImage()
{
element=document.getElementById('pokemon_card')
if
(element.src.match("http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card- back.jpg?w=750"))
{element.src="'.$result['image_url'].'";} //<- no idea how to express the php string variable here
else
{element.src="http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card-back.jpg? w=750";}
}
</script>
</head>
<body>
<?php
$dbhost = 'databasePlace';
$dbname = 'mine';
$dbuser = 'me';
$dbpass = '******';
$link = mysqli_connect($dbhost,$dbuser,$dbpass,$dbname);
mysqli_select_db($link,$dbname);
$name = $_GET["fname"];
$query = sprintf("SELECT image_url, Type
FROM Pokemon c
WHERE c.name='%s'",
mysqli_real_escape_string($link,$name));
$result = mysqli_fetch_assoc(mysqli_query($link,$query));
echo '<img id="pokemon_card" onclick="changeImage()" height="225"
width="165" src="'.$result['image_url'].'"/>';
mysqli_close($link);
?> …