mysqli SELECT查询不起作用,原因不明
问题描述:
这是我的php连接数据库和查询表“userActivityTime”,其中有一行。我没有连接到数据库(即没有错误)的麻烦,但我的查询不起作用,尽管看遍了整个互联网,我无法弄清楚原因。希望你们都能帮忙。非常感谢!mysqli SELECT查询不起作用,原因不明
<?php
// ESTABLISH TABLE AND COLUMN NAMES
$mysqli = new mysqli("*****", "****", "*****", "****");
// MAKE SURE CONNECTION SUCCEEDED
if ($mysqli_connection->connect_error) {
echo "Not connected, error: " . $mysqli_connection->connect_error;
exit();
} else {
echo "connected";
}
$query = "SELECT 'userDailyTime' FROM 'userActivityTime'";
if ($mysqli->query($query)) {
echo $mysqli->error;
}
exit();
?>
答
采用背蜱如果必要的话不是单引号:
SELECT `userDailyTime` FROM `userActivityTime`
或只是
SELECT userDailyTime FROM userActivityTime
编辑:
从互联网上的一个例子,蠕虫你的东西融入概念。
<?php
$servername = "localhost";
$username = "username";
$password = "password";
$dbname = "myDB";
// Create connection
$conn = new mysqli($servername, $username, $password, $dbname);
// Check connection
if ($conn->connect_error) {
die("Connection failed: " . $conn->connect_error);
}
$sql = "SELECT id, firstname, lastname FROM MyGuests";
$result = $conn->query($sql);
if ($result->num_rows > 0) {
// output data of each row
while($row = $result->fetch_assoc()) {
echo "id: " . $row["id"]. " - Name: " . $row["firstname"]. " " . $row["lastname"]. "<br>";
}
} else {
echo "0 results";
}
$conn->close();
?>
答
您不需要将表名和列名放在引号中。当查询成功时,您正在告诉它回显$mysqli->error
(用一个感叹号标记它)。此外,您创建$mysqli
然后将其称为$mysqli_connection
,所以我不知道您的代码如何工作。
答
试试这个:
<?php
// ESTABLISH TABLE AND COLUMN NAMES
$mysqli = new mysqli("*****", "****", "*****", "****");
// MAKE SURE CONNECTION SUCCEEDED
if ($mysqli->connect_errno) {
echo "Not connected, error: " . $mysqli->connect_error;
return false;
} else {
echo "connected";
}
$query = "SELECT userDailyTime FROM userActivityTime";
if (!$mysqli->query($query)) {
echo $mysqli->error;
}
return true;
?>
如果你真的有没有错误,那么这段代码就什么也不做,所以这将是不可能为“不行”。你能否将你的问题更清楚地说明你得到了什么结果? –
'userActivityTime'是一个字符串 – Strawberry