2015-07-19 94 views
-4

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(); 

?> 
+0

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? –

+0

'userActivityTime'是一个字符串 – Strawberry

回答

2

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(); 
?> 
+3

OP也使用了错误的变量'$ mysqli_connection' –

+0

dang Fred错过了ya,半个月 – Drew

+0

我试过了,没有引号,即使那样也行不通。 –

1

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。

1

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; 

?>