PHP / MySqli:使用2个表准备MySqli语句,并且WHERE不返回数据

时间:2015-11-14 16:00:29

标签: php xml mysqli

我正在尝试从我的罪犯数据库的2个表中获取数据,并且当我添加" AND criminal_name =' FIDEL URBINA'"只获取每个犯罪分子的数据。

这是我现在的代码和SQL语句,

$stmt = $mysqli->prepare("SELECT criminal_name, event_location FROM criminal_profile, criminal_locations 
    WHERE criminal_profile.criminal_id = criminal_locations.criminal_id");

$stmt->execute();
$stmt->bind_result($criminal_name, $event_location); //get data from statement
$stmt->store_result();


 if($formaat=='xml'){ //XML
            $xml=new SimpleXMLElement('<criminals></criminals>');


            while($stmt->fetch()) // zolang er rijen zijn
            {
                // $criminal_name, Scriminal_pob en $event_location zijn gevuld
                //voeg element criminal_pob toe met naam criminal_pob, criminal_name en event_location:
                $info = $xml->addChild('criminal');
                $item = $info->addChild('criminal_name',$criminal_name);
                $item = $info->addChild('event_location',$event_location);
            }
            header('Content-type: text/xml'); //DIT WERKT NIET? cant send headers after they were sent

            // coderen als JSON:
            echo $xml->asXML();
        }

根据他们拥有的事件位置重复犯罪分子,如下图所示:

http://grabilla.com/05b0e-a271d885-dc45-47ad-a043-c92d832162bd.png

任何人都知道为什么在我添加&#34; AND criminal_name =&#39; FIDEL URBINA&#39;&#34;或者类似的东西?我也试过&#34; AND criminal_name =&#39; F%&#39;&#34;。

1 个答案:

答案 0 :(得分:1)

可能是匹配问题,您可以尝试使用templates/<app_name>操作员而不是like

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