无法从MySQL表检索与特定“ uid”匹配的数据

无法从MySQL表检索与特定“ uid”匹配的数据

在您的Android应用中，您需要一个JSONArray：

// store incoming stream in an arrayJSonArray jArray = new JSonArray(streamToString(instream));

但是，在您的PHP文件中，您仅输出多个单独的JSON对象，而不是实际数组。我认为，您应该首先从数据库中的PHP数组中收集所有项目，然后再对其进行编码和输出一次。

我的PHP技能有点生锈，但是我希望这能起作用：

//store # of rows returned$num_rows = mysql_num_rows($query);if ($num_rows >= 1) {    $output = array();    while($results = mysql_fetch_assoc($query)) {        // append row to output        $output[] = results    }    mysql_close();  // shouldn't that be outside the if block?    //enpre the returned data in JSON format    echo json_enpre($output);}

我希望输出是这样的（也许没有缩进）：

[    {"nid":"1","vid":"1","type":"goal","language":"","title":"test","uid":"1","status":"1","created":"1342894493","changed":"1342894493","comment":"2","promote":"1","moderate":"0","sticky":"1","tnid":"0","translate":"0"},    {"nid":"2","vid":"2","type":"goal","language":"","title":"test2","uid":"1","status":"1","created":"1342894529","changed":"1342894529","comment":"2","promote":"1","moderate":"0","sticky":"1","tnid":"0","translate":"0"},    {"nid":"5","vid":"5","type":"goal","language":"","title":"run","uid":"1","status":"1","created":"1343506987","changed":"1343506987","comment":"2","promote":"1","moderate":"0","sticky":"1","tnid":"0","translate":"0"},    {"nid":"9","vid":"9","type":"goal","language":"","title":"run to the hills","uid":"1","status":"1","created":"1343604338","changed":"1343605100","comment":"2","promote":"0","moderate":"0","sticky":"0","tnid":"0","translate":"0"}]