php 在非对象上调用成员函数 execute()
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call to a member function execute() on a non-object
提问by Malloc
My script containing that error is this:
我的包含该错误的脚本是这样的：
$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
在服务器（不是本地主机）上运行的 php 版本是 5.2.17
回答by Pierre de LESPINAY
$stmtis supposed to be an object with the method execute().
Seems like $this->db->prepare()is not returning the good result.
$stmt应该是具有方法的对象execute()。
似乎$this->db->prepare()没有返回好结果。
If $this->dbis a mysqli()object you should bind the parameterslike that:
如果$this->db是一个mysqli()对象，你应该像这样绑定参数：
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();
// ...
}
回答by xdazz
Check the sql you executed,
检查你执行的sql，
$this->db->prepare('SELECT libelle,activite,adresse,tel,lat,lng FROM etablissements where type IN ('.$in_list.')');
does not return a valid statement object.
不返回有效的语句对象。
回答by suraj kumar
Swap the statement bind and execute and replace result with param, It will work fine
交换语句绑定并执行并用参数替换结果，它会正常工作
$stmt = $this->db->prepare('SELECT libelle,activite,adresse,tel,lat,lng FROM etablissements
where type IN ('.$in_list.')');
$stmt->bind_param($libelle,$activite,$adresse,$tel,$lat,$lng);
$stmt->execute();
回答by nz-io
Your db object is null. Check if you close the connection too early or somehow you override it to null.
您的 db 对象为空。检查您是否过早关闭连接或以某种方式将其覆盖为空。
