我正试图通过在url中传递参数来从JSON返回特定数据，示例如下。
http://localhost/api/api.php?post_title=Strawbrerry
这绝对没有任何作用，希望能就如何解决这一问题提供一些建议。。。。请参阅下面的代码。
$connection = @mysqli_connect($server, $user, $password, $db);
if( ! $connection ) die( "Error ".mysqli_connect_error() );
$sql = "SELECT * FROM posts";
$result = mysqli_query($connection, $sql);
$array_post = array();
while($data = mysqli_fetch_assoc($result)){
$array_post['post_title'][] = $data['post_title'];
$array_post['post_description'][] = $data['post_description'];
$array_post['post_image'][] = $data['post_image'];
$array_post['posted_by]'][] = $data['posted_by'];
$array_post['[post_date]'][] = $data['post_date'];
}
echo json_encode($array_post);
如果您想依靠名为post_title的URL参数来返回标题中包含单词"Strawberry"的发布结果，您可以使用以下行修改当前脚本：
$sql = "SELECT * FROM posts";
if(isset($_GET['post_title']) && $_GET['post_title']) {
$sql .= " WHERE post_title LIKE '%?%'";
// Use prepared statements here. Don't trust GET parameters
// to not be SQL injection attempts.
$stmt = mysqli_prepare($connection, $query);
mysqli_stmt_bind_param($stmt, 's', $_GET['post_title']);
mysqli_stmt_execute($stmt);
mysqli_stmt_bind_result($stmt, $result);
} else {
$result = mysqli_query($connection, $sql);
}
// Here, $result should hold the result set
如果未在URL中设置post_title，则不会发生任何更改。如果是，它会修改您正在构建的SQL语句，使其包含LIKE表达式，以过滤查询结果，使其仅包含标题包含通过URL传入的内容的结果。有关更多信息，请参阅以下手册条目。
mysqli_preparemysqli_stmt_bind_parammysqli_stmt_executemysqli_stmt_bind_result