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ajax php传递和接收变量实现思路及代码_PHP教程

程序员文章站 2024-02-14 08:30:10
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So, your jQuery might be something like.....
复制代码 代码如下:

$.ajax({
url: 'query.php',
data: {id:10},
datatype: json
success: function(results) {
if (results.msg == 'success') {
for (var i in data) {
$('#content').append(
'id = ' + results.data[i].id + ', description = ' + results.data[i].description + ', msrp = ' + results.data[i].msrp
);
}
} else {
$('#content').append(results.msg);
}
}
});

And your php....
复制代码 代码如下:

if (isset($_GET['id'])) {
$sql = "SELECT id, description, msrp FROM tbl WHERE id = '{$_GET['id']}'";
$return = array();
if ($result = mysql_query($sql)) {
if (mysql_num_rows($result)) {
$return['msg'] = 'success';
while ($row = mysql_fetch_assoc($result)) {
$return['data'][] = $row;
}
} else {
$return['msg'] = 'No results found';
} else {
$return['msg'] = 'Query failed';
}
header("Content-type: application/json");
echo json_encode($result);
}

www.bkjia.comtruehttp://www.bkjia.com/PHPjc/326435.htmlTechArticleSo, your jQuery might be something like..... 复制代码 代码如下: $.ajax({ url: 'query.php', data: {id:10}, datatype: json success: function(results) { if (results.msg == 'suc...