PHP代码没有得到MySQL表值

我已经看到了很多像这样的问题，但没有一个真正帮助我。我试图编写一些从MySQL表中获取值并将其打印出来的代码。每当我尝试运行代码时，它都不会运行它，只是显示原始代码，或者它返回一个MySQL错误。我的PHP如下：PHP代码没有得到MySQL表值

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

$con = mysqli_connect($servername, $username, $password, $dbname); 


$sql = "SELECT * FROM `class1`"; 


$result = mysql_query($sql, $con); 


if (!$result) { 
    echo "Could not successfully run query ($sql) from DB: " . mysql_error(); 
    exit; 
} 


$data = ""; 
while ($row = mysql_fetch_assoc($result)) { 
    $data .= "<tr> 
       <td>" . $row['name'] . "</td> 
       <td>" . $row['score'] . "</td> 


       </tr>"; 
} 

exit; 

?>

顺便说一句，该错误信息是'无法成功运行查询（SELECT * FROM从DB的Class1：”

我知道这个问题很普遍，所以。请告诉我，如果我需要添加任何东西！

来源

2015-02-23 javathunderman

为什么混淆了'mysqli_connect'和'mysql_'函数？当然这不会起作用。 – bcesars 2015-02-23 20:41:19

（facepalm）对。我在PHP很新，你能告诉我如何解决这个问题吗？ – javathunderman 2015-02-23 20:42:08

确保不要使用'mysql_'，而是'mysqli_' ... – developerwjk 2015-02-23 20:42:57

$con = mysqli_connect($servername, $username, $password, $dbname); 
      ^----mysql **I** 

$result = mysql_query($sql, $con); 
       ^---mysql **WITHOUT** I

如果你有绝对的连骨头裸露在你的代码最小的错误处理，而不是盲目地假设没有什么能出问题，你会被告知，你的查询失败，因为没有连接到数据库：

$result = mysql_query($sql, $con) or die(mysql_error()); 
           ^^^^^^^^^^^^^^^^^^^^^^

来源

2015-02-23 20:48:01

你介意发布固定代码吗？谢谢。 – javathunderman 2015-02-23 21:09:36

你不能在自己的代码中将'mysql'更改为'mysqli'？如果你太忙，无法几次击中'i'键，那么你太忙了，无法在这个网站上... – 2015-02-23 21:10:28

不知道有没有...不好意思。 – javathunderman 2015-02-23 21:23:28

基本上，它看起来像您正在使用的mysqli混合mysql的

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

$con = mysqli_connect($servername, $username, $password, $dbname); 


$sql = "SELECT * FROM `class1`"; 


$result = mysqli_query($con,$sql) or die(mysqli_error($con)); 



$data = ""; 
while ($row = mysqli_fetch_assoc($result)) { 
    $data .= "<tr> 
       <td>" . $row['name'] . "</td> 
       <td>" . $row['score'] . "</td> 


       </tr>"; 
} 

exit(); 

?>

来源

2015-02-23 21:13:03 brad

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

$con = mysqli_connect($servername, $username, $password, $dbname); 


$sql = "SELECT * FROM `class1`"; 


$result = mysqli_query($sql, $con); 


if (!$result) { 
    echo "Could not successfully run query ($sql) from DB: " .  mysql_error(); 
    exit; 
    } 


$data = ""; 
while ($row = mysqli_fetch_assoc($result)) { 
    $data .= "<tr> 
       <td>" . $row['name'] . "</td> 
       <td>" . $row['score'] . "</td> 


      </tr>"; 
} 

exit; 

?>

您使用的MySQLi用于连接到数据库，所以你必须在功能

使用 “库MySQLi”，而不是 “MySQL的”

来源

2015-02-23 21:29:01 zheek

嗨Java迅雷侠，

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

    $con = mysql_connect($servername, $username, $password, $dbname); 
    //---Make sure those variables for connecting DB are correct 
    //And please using one DB connection method (PDO/mysql/mysqli so on) 

    $sql = mysql_query("SELECT * FROM `class1`") or die(mysql_error()); 

    $data = "<html>"; 
    $data .= "<body>"; 
     $data .= "<table>"; 
     if(mysql_num_rows($sql) > 0) {//list the result if rows of $sql > 0 
       while ($rows = mysqli_fetch_assoc($sql)) { 
        $data .= "<tr>"; 
        $data .= "<td>" . $row['name'] . "</td>"; 
        $data .= "<td>" . $row['score'] . "</td>"; 
        $data .= "</tr>"; 
       } 
     } 
     $data .= "</table>"; 
    $data .= "</body>"; 
    $data .= "</html>"; 
    echo $data; //Show the results by using echo 

?>

我也是PHP中的新成员，我不确定在哪条线上出现错误，除非您没有回显$数据。此外，你可以更清楚地说明错误吗？谢谢。

来源

2015-02-24 09:17:32

运行此操作，它应该适用于您，如果您尝试查询您的表和列名称中没有拼写错误。

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

$con = new mysqli($servername, $username, $password, $dbname); 

$sql = "SELECT * FROM class1"; 

$result = $con->query($sql); 

if ($con->error) { 

    die("Could not connect to database"); 

} else { 

    $data = ""; 

    while ($row = mysqli_fetch_assoc($result)) { 

     $data .= "<tr> 
        <td>" . $row['name'] . "</td> 
        <td>" . $row['score'] . "</td> 
        </tr>"; 

    } 

    echo $data; 
    exit; 
} 

?>

来源

2015-02-24 18:42:30

PHP代码没有得到MySQL表值

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