php mysqli_result 类的对象无法转换为字符串
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Object of class mysqli_result could not be converted to string
提问by Joann
I asked Google to help me I got no luck. :-( Here's the particular code that generates the error:
我问谷歌帮助我我没有运气。:-( 这是产生错误的特定代码：
$this->conn->query("UPDATE tz_members SET confirm='yes' WHERE usr='".$uname."'");
The whole function is the following:
整个函数如下：
function update_confirm_field($code) {
$uname = $this->conn->query("SELECT usr FROM tz_members WHERE
confirm='".$code."'");
$this->conn->query("UPDATE tz_members SET confirm='yes' WHERE
usr='".$uname."'");
}
Forgive me if I have missed something stupid. Can anyone tell me what's causing the problem please???
如果我错过了一些愚蠢的事情，请原谅我。谁能告诉我是什么导致了问题？？？
回答by wash
The problem is that $uname is an object, not a string. You need to call one of $uname's methods to access the data.
问题是 $uname 是一个对象，而不是一个字符串。您需要调用 $uname 的方法之一来访问数据。
function update_confirm_field($code) {
$uname = $this->conn->query("SELECT usr FROM tz_members WHERE
confirm='".$code."'");
while ($row = $uname->fetch_assoc()) {
$this->conn->query("UPDATE tz_members SET confirm='yes' WHERE
usr='".$row["usr"]."'");
}
}
that should do it (or one of the above solutions).
应该这样做（或上述解决方案之一）。
回答by Luis Melgratti
$unamereturned by your first query is a mysql_result object, not a string. you must fetch the data from that result in order to use it in your second query.
您的第一个查询返回的$uname是一个 mysql_result 对象，而不是一个字符串。您必须从该结果中获取数据才能在第二个查询中使用它。
while ($row = mysql_fetch_assoc($result)) {
echo $row["usr"];
}
回答by Mitch Dempsey
The querymethod returns a pointer/object of the query result, it does not just directly dump the response. You need to do something like list($uname) = $uname->fetch_row;
该query方法返回查询结果的指针/对象，它不只是直接转储响应。你需要做类似的事情list($uname) = $uname->fetch_row;
回答by Phill Pafford
$updateQuery = "UPDATE tz_members SET confirm='yes' WHERE usr= (SELECT usr FROM tz_members WHERE confirm='".$code."')";
// Get name and update in the same query
$this->conn->query($updateQuery);
