我不知道JS位置可能会改变我的html页面的reults,直到今天。我希望图像src在两次不同的网址之间切换。为什么这个第一个代码像我想要的那样工作,但第二个代码不工作?第二个代码的源html为var not_a_bad_word生成一个空白字符串。为什么我的JS的位置会改变我的html代码?
第一代码:
'Murica!'$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 '
width="330" src="' . $result['image_url'] . '"/>';
mysqli_close($link);
?>
function changeImage() {
element = document.getElementById('pokemon_card');
var not_a_bad_word = "<?php echo $result['image_url']; ?>";
if (element.src == "http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card-back.jpg") {
element.src = not_a_bad_word;
}
else {
element.src="http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card-back.jpg";
}
}
第二代码:
'Murica!function changeImage() {
element = document.getElementById('pokemon_card');
var not_a_bad_word = "<?php echo $result['image_url']; ?>";
if (element.src == "http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card-back.jpg") {
element.src = not_a_bad_word;
}
else {
element.src="http://dmisasi.files.wordpress.com/2010/12/david-pokemon-card-back.jpg";
}
}
$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 '
width="330" src="' . $result['image_url'] . '"/>';
mysqli_close($link);
?>
+0
而控制台说什么? –
+0
当您向其他人展示您的代码时,请格式化为好。代码只写入一次,但多次读取,因此值得额外付出努力(对同行程序员来说也是常见的礼节)。我已经在你的问题中格式化了代码。 –