2015-07-19 42 views
-4

這是我的php連接數據庫和查詢表「userActivityTime」,其中有一行。我沒有連接到數據庫(即沒有錯誤)的麻煩,但我的查詢不起作用,儘管看遍了整個互聯網,我無法弄清楚原因。希望你們都能幫忙。非常感謝!mysqli SELECT查詢不起作用,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

如果你真的有沒有錯誤,那麼這段代碼就什麼也不做,所以這將是不可能爲「不行」。你能否將你的問題更清楚地說明你得到了什麼結果? –

+0

'userActivityTime'是一個字符串 – Strawberry

回答

2

採用背蜱如果必要的話不是單引號:

SELECT `userDailyTime` FROM `userActivityTime` 

或只是

SELECT userDailyTime FROM userActivityTime 

編輯:

從互聯網上的一個例子,蠕蟲你的東西融入概念。

<?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也使用了錯誤的變量'$ mysqli_connection' –

+0

dang Fred錯過了ya,半個月 – Drew

+0

我試過了,沒有引號,即使那樣也行不通。 –

1

您不需要將表名和列名放在引號中。當查詢成功時,您正在告訴它回顯$mysqli->error(用一個感嘆號標記它)。此外,您創建$mysqli然後將其稱爲$mysqli_connection,所以我不知道您的代碼如何工作。

1

試試這個:

<?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; 

?>