PHP和MySQL連接不工作

Question

我已經寫了下面的代碼來連接MySQL中的PHP，但我沒有得到輸出。

<?php 
    $servername = "localhost"; 
    $username = "root"; 
    $password = "pravin"; 
    $mysql_conn = new mysqli($servername, $username, $password); 
     if ($mysql_conn->connect_error) { 
      die("Connection failed: ". $mysql_conn->connect_error); 
} 

echo "Connected successfully"; 
$name = $_POST["microorganism"]; 
echo $name; 
$db_selected = mysql_select_db('yieldofvanillin', $mysql_conn); 
if (!$db_selected){ 
    die ('Can\'t use : ' . mysql_error()); 
} 
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' "; 
$result = $mysql_query($query); 
while ($line = myql_fetch_array($result, MYSQL_ASSOC)) { 
    echo $line["Substrate"]; 
    echo $line["products"]; 
    echo $line["Microorganism"]; 
    echo $line["yield"]; 
    echo $line["Reference"]; 
} 

mysql_close($mysql_conn); 
?> 

數據庫名稱是「yieldofvanillin」，它有五列。我獲得輸出連接成功。之後，沒有輸出。請讓我知道代碼中的錯誤。

Answer 1

我已刪除錯誤。我在評論中提到。代碼參考PHP Manual。你應該閱讀本手冊（強烈推薦）

<?php 
$mysqli = new mysqli("localhost", "root", "pravin", "yieldofvanillin"); 

/* check connection */ 
if ($mysqli->connect_errno) { 
    printf("Connect failed: %s\n", $mysqli->connect_error); 
    exit(); 
} 

$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' "; 

if ($result = $mysqli->query($query)) { 

    /* fetch associative array */ 
    while ($row = $result->fetch_assoc()) { 
     echo $row["Substrate"]; 
     echo $row["products"]; 
     echo $row["Microorganism"]; 
     echo $row["yield"]; 
     echo $row["Reference"]; 

    } 

    /* free result set */ 
    $result->free(); 
} 

Answer 2

你在混合mysqli和mysql庫。

的代碼應該是：在你的code.Read

<?php 
$servername = "localhost"; 
$username = "root"; 
$password = "pravin"; 
$mysql_conn = new mysqli($servername, $username, $password); 
    if (mysqli_connect_errno()) { 
     die("Connection failed: ". mysqli_connect_error()); 
} 

echo "Connected successfully"; 
$name = $_POST["microorganism"]; 
echo $name; 
$db_selected = mysqli_select_db($mysql_conn,'yieldofvanillin'); 
if (!$db_selected){ 
    die ('Can\'t use : ' . mysqli_error($mysql_conn)); 
} 
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' "; 
$result = mysqli_query($mysql_conn,$query); 
while ($line = mysqli_fetch_assoc($result)) { 
    echo $line["Substrate"]; 
    echo $line["products"]; 
    echo $line["Microorganism"]; 
    echo $line["yield"]; 
    echo $line["Reference"]; 
} 

mysqli_close($mysql_conn); 
?> 

Answer 3

刪除錯誤仔細PHP手冊。

<?php 
     $servername = "localhost"; 
     $username = "root"; 
     $password = "pravin"; 
     $db = "yieldofvanillin"; 

    // Create connection 
     $mysqli = new mysqli($servername, $username, $password, $db); 

    /* connection string*/ 
    if ($mysqli->connect_errno) { 
     die("Connection failed: " . $mysqli->connect_error); 
     exit(); 
    } 

    $query = "SELECT * FROM vanillin WHERE Microorganism = '$name' "; 

    if ($result = $mysqli->query($query)) { 


    while ($row = $result->fetch_assoc()) { 
      echo $row["Substrate"]; 

     } 


    $result->free(); 
    } 

$mysqli->close(); 

？>

你的輸出，因此無法顯示mysql_fetch_array不correct.Because你混合mysql_和mysqli_功能，你叫myql_fetch_array中不存在mysqli的。 MySQL和MySQLi是兩種不同的PHP擴展，它們不能混用。因爲前者在mysqli