ajax php传递和接收变量实现思路及代码_PHP教程

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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);

}

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  • 本文由 发表于 2020年7月7日 03:02:01
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