PHP not displaying correctly in HTML
I'm trying to create a drop down box that is populated by an array. When I view my code the browser just displays pieces of the php code.
Here is what the browser displays:
'; foreach ($array as $value) { $html .= '
$value
'; } $html .= ''; return $html; } ?>
Here is my code:
<html>
<head>
<title>test</title>
<?php
$cities = array( 'San Francisco', 'San Diego', 'Los Angeles');
function createDropDown($name, $array) {
$html = '<select name=$name>';
foreach ($array as $value) {
$html .= '<option value="$value">$value</option>';
}
$html .=开发者_StackOverflow '</select>';
return $html;
}
?>
</head>
<body>
<form>
<?php
createDropDown("cities", $cities);
?>
</form>
</body>
</html>
You need to echo your output as well. You build the string but never output.
<html>
<head>
<title>test</title>
<?php
$cities = array( 'San Francisco', 'San Diego', 'Los Angeles');
function createDropDown($name, $array) {
$html = '<select name="' . $name . '">';
foreach ($array as $value) {
$html .= '<option value="' . $value . '">' . $value . '</option>';
}
$html .= '</select>';
return $html;
}
?>
</head>
<body>
<form>
<?php
echo createDropDown("cities", $cities);
?>
</form>
</body>
</html>
Most likely your server isn't configured to treat .php files as PHP scripts, and is serving them up as html/plaintext. If you view-source the page this is happening on, you'll probably get the full PHP source code. The <?php is seen as an opening HTML tag, which isn't closed until just before the foreach loop, so all the intermediate text/code is "hidden" in HTML view.
Change
$html = '<select name=$name>';
to
$html = '<select name="' . $name. '">';
Change
$html .= '<option value="$value">$value</option>';
to
$html .= '<option value="' . $value . '">' . $value. '</option>';
' doesn't parse, but " does.
$html = "<select name=$name>";
will do it.
加载中,请稍侯......
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