php不與mysql連接

-1

我是新來的網頁設計，我使用PHP和MAMP服務器的崇高文本編輯器。當我在瀏覽器中運行我的頁面時，我認爲MYSQL數據庫已連接，但沒有得到正確的輸出。我PGM低於：php不與mysql連接

basicform.html

<html> 
<head> 
<title>A BASIC HTML FORM</title> 
</head> 
<body> 
    <Form name ="form1" Method ="POST" Action ="submitForm.php"> 

     <INPUT TYPE = "TEXT" VALUE ="username" Name ="username"> 
     <INPUT TYPE = "Submit" Name = "Submit1" VALUE = "Login"> 
    </FORM> 
</body> 
</html>

submiteForm.php

<?PHP 
$con = mysql_connect("localhost","root","root"); 

if (mysqli_connect_errno($con)) { 
    echo "Failed to connect to MySQL: " . mysqli_connect_error(); 
} else echo "success"; 

mysql_select_db("innowell",$con) or die(mysql_error()); 

$query = "SELECT * FROM employees"; 
$result = mysql_query($query); 

echo $result; 

$username = $_POST['username']; 

if ($username == "letmein") { 
    print ("Welcome back, friend!"); 
} else { 
    print ("You're not a member of this site"); 
} 

?>

輸出：

成功的資源ID＃3你是不是該成員現場。

我的pgm.can中有什麼錯誤有人幫我解決問題嗎？

來源

2014-09-19 anu l

你不取結果。 – 2014-09-19 10:12:50

你在混合MySQL API。 – 2014-09-19 10:13:17

恭喜，輸出證明你的代碼工作得很好。 – 2014-09-19 10:14:13

您目前正在混合使用MySQL API，它們（mysql_和mysqli_）功能不會混在一起。

使用像這樣的，拿上mysqli_connect()

//conection: 
$link = mysqli_connect("myhost","myuser","mypassw","mydb") or die("Error " . mysqli_error($link)); 

//consultation: 

$query = "SELECT name FROM mytable" or die("Error in the consult.." . mysqli_error($link)); 

$result = mysqli_query($link, $query); 

//display information: 

while($row = mysqli_fetch_array($result)) { 
    echo $row["name"] . "<br>"; 
}

來源

2014-09-19 10:15:07

$con=mysqli_connect('localhost','root','','innowell') or die(mysqli_error($con)); 

$query = "SELECT name FROM mytable" or die("Error in the consult.." . mysqli_error($con)); 

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

//display information: 

while($row = mysqli_fetch_array($result)) { 
    echo $row["ColumnName"] . "<br>"; 
}

來源

2014-09-19 10:15:26

有一些東西在答案錯誤，請讓我知道謝謝 – 2014-09-19 10:17:42

OP是混合的MySQL APIs。你沒有做出正確的改變。 – 2014-09-19 10:18:35

@ Fred-ii非常感謝。我刪除它是錯誤的。我沒有注意到我的壞之前謝謝你回答。 – 2014-09-19 10:19:44

看看

要麼用mysql或mysqli的

和更新象下面這樣：

  $db = mysql_select_db("innowell",$con) or die(mysql_error()); 
      $query = "SELECT name FROM employees"; 
      $result = mysql_query($query,$db); 
      while($row = mysql_fetch_array($result)) { 
       echo $name = $row["name"]; 
      }

來源

2014-09-19 10:34:12 viveik10

-----basicform.html---- 
    <html> 
    <head> 
    <title>A BASIC HTML FORM</title> 
    </head> 
    <body> 

    <form name="form1" method="POST" action ="submitForm.php"> 

    <input type="text" value="username" name="username" /> 
    <input type="submit" name = "submit1" value="Login" /> 

    </form> 

    </body> 
    </html> 

        \\submitForm.php\\ 

    <?php 
$con = mysql_connect("localhost", "root", "root"); 
if (!$con) { 
    echo "Failed to connect to MySQL: " . mysql_error(); 
} else { 
    echo "success"; 
} 
mysql_select_db("innowell", $con) or die(mysql_error()); 
$username = $_POST['username']; 
$query = "SELECT * FROM employees where username='" . $username . "'"; 
$result = mysql_query($query); 


if (mysql_num_rows($result)) { 

    print("Welcome back, friend!"); 

} else { 

    print("You're not a member of this site"); 

} 

?>

來源

2014-09-19 10:51:14

我假設你在僱員表中有'用戶名'字段。 – 2014-09-19 10:52:02

php不與mysql連接

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