Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
386 views
in Technique[技术] by (71.8m points)

php - call to a member function execute() on a non-object

My script containing that error is this:

$stmt = $this->db->prepare('SELECT libelle,activite,adresse,tel,lat,lng FROM etablissements where type IN ('.$in_list.')');
        $stmt->execute();
        $stmt->bind_result($libelle,$activite,$adresse,$tel,$lat,$lng);

The php version running on the server (not localhost) is 5.2.17

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

$stmt is supposed to be an object with the method execute().
Seems like $this->db->prepare() is not returning the good result.

If $this->db is a mysqli() object you should bind the parameters like that:

if ($stmt = $this->db->prepare('SELECT libelle,activite,adresse,tel,lat,lng FROM etablissements where type IN (?)')) {
  $stmt->bind_param("s", $in_list);
  $stmt->execute();
  // ...
}

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...