PHP Simple implementation Query database return JSON data, PHP database returned json_php tutorial

PHP Simple implementation Query database return JSON data, PHP database return JSON

Example code one:

Sets the return JSON format Data header (' Content-type:application/json;charset=utf8 ');//Connect Database $link = mysql_connect ("localhost", " Root "," root ") or Die (" unable-to-connect to the mysql! "); mysql_query ("SET NAMES ' UTF8 '"); mysql_select_db ("Jilinwula", $link) or die ("Unable to connect to the mysql!"); /Get paging parameter $page = 0; $pageSize = 3;if (!is_null ($_get["page")) {$page = $_get["page"];} if (!is_null ($_get["PageSize")) {$pageSize = $_get["PageSize"];} Query the data into the array $result = mysql_query ("Select Username,password from UserInfo limit". $page. ", ". $pageSize. ""); $results = Array (), while ($row = Mysql_fetch_assoc ($result)) {$results [] = $row;} Turn the array into JSON format echo json_encode ($results);//close Connection mysql_free_result ($result); Mysql_close ($link);

Example code two:

<?php//need to execute SQL statement//single $sql= "Select Id,name from Tbl_user where id=1";//Multiple data//$sql = "Select Id,name from Tbl_user";// Call the conn.php file for database operation require (' conn.php '); Prompt Operation success Information, note: $result exists in the conn.php file, is called out if ($result) {//$array =mysql_fetch_array ($result, MYSQL_ASSOC);   /* Data set $users =array (); $i = 0; while ($row =mysql_fetch_array ($result, Mysql_assoc)) {  echo $row [' id ']. ' -----------'. $row [' name ']. '
';  $users [$i]= $row;  $i + +; } Echo Json_encode (Array (' dataList ' = $users)); */* Single data */$row =mysql_fetch_row ($result, MYSQL_ASSOC);  echo json_encode (Array (' jsonobj ' = $row));} Mysql_free_result ($result);//release result mysql_close ();//close Connection?>

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