2017-07-05 125 views
0

我有一個關於phpmyAdmin的數據庫,它由三列組成:id,名稱和編號。 我已經通過phpmyadmin向數據庫添加了3行數據。 我現在想通過我的php文件將數據添加到這個數據庫。這是我使用的數據添加,並顯示在瀏覽器上的數據代碼:SQL錯誤(錯誤:無法執行INSERT INTO)

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = "root"; 
$dbname = "myfirstsite"; 

// Create connection 
$conn = new mysqli($servername, $username, $password, $dbname); 
// Check connection 
if ($conn->connect_error) { 
die("Connection failed: " . $conn->connect_error); 
} else { 
echo "wooo connected"; 
} 

$sql = "INSERT INTO hi (id, name, number) 
VALUES ('99', 'Doe', '999999')"; 
if(mysqli_query($link, $sql)){ 
echo "Records inserted successfully."; 
} else{ 
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); 
} 


//displaying data 
$sql = "SELECT id, name, number FROM hi"; 
$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["name"]. " " .     $row["number"]. "<br>"; 
} 
} else { 
echo "0 results"; 
} 

$conn->close(); 
?> 

的事情是,我不明白爲什麼新的數據不放入數據庫,但目前數據顯示在屏幕上。

回答

2

在你的代碼是$link其犯規存在做參考,它應該是$conn

Chaneg這樣:

if(mysqli_query($link, $sql)){ 
echo "Records inserted successfully."; 
} else{ 
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); 
} 

要:

if(mysqli_query($conn, $sql)){ 
echo "Records inserted successfully."; 
} else{ 
echo "ERROR: Could not able to execute $sql. " . mysqli_error($link); 
} 
+0

它完美現在,謝謝。 – pete800