無法連接到數據庫？

Question

這是我下面的php代碼。 出於某種原因，當我希望它連接到數據庫時，它會顯示“沒有選擇數據庫”，但是當我進行連接測試時，它會顯示“數據庫連接成功！” 因此，下面的代碼中存在錯誤。 另外我認為它可能會干擾我的形式，所以我必須輸錯字; 當我刪除PHP時，表單正常運行，但我希望它連接到數據庫。

<?php
// let's initialize vars to be printed to page in the HTML section so our script does not return errors 
// they must be initialized in some server environments, not shown in video
$errorMsg = "";
$First_Name = "";
$Last_Name = "";
$Email = "";
$Password = "";


// This code runs only if the form submit button is pressed
if (isset ($_POST['firstname'])){

    /* Example of cleaning variables in a loop
    $vars = "";
    foreach ($_POST as $key => $value) {
       $value = stripslashes($value);
       $vars .= "$key = $value<br />";
    }
    print "$vars";
    exit();
    */
     $First_Name = $_POST['firstname'];
     $Last_Name = $_POST['lastname'];
     $Email = $_POST['email'];
     $Password = $_POST['password'];

     // Connect to database
     include "connect_to_mysql.php";
     $emailCHecker = mysql_real_escape_string($Email);
     $emailCHecker = eregi_replace("`", "", $EmailCHecker);
     // Database duplicate e-mail check setup for use below in the error handling if else conditionals
     $sql_email_check = mysql_query("SELECT email FROM members WHERE email='$emailCHecker'");
     $email_check = mysql_num_rows($sql_email_check); 

}

     $sql = mysql_query("INSERT INTO members (firstname, lastname, email, password, sign_up_date) 
     VALUES('$First_Name','$Last_Name','$Email','$Password', now())")  
     or die (mysql_error());

     $id = mysql_insert_id();

     // Create directory(folder) to hold each user's files(pics, MP3s, etc.)        
     mkdir("members/$id", 0755);    

?>

這是我在下面的表格編碼

    <form action="index.php" method="post" enctype="multipart/form-data">
  <tr>
    <td width="23%" class="right">First Name:</td>
    <td width="77%" class="left left_nowrap"><input type="text" class="left left_nowrap tb10" id="First_Name" value="<?php print "$First_Name"; ?>"/></td>
  </tr>
  <tr>
    <td class="right">Last Name:</td>
    <td class="left left_nowrap"><input type="text" class="left left_nowrap tb10" id="Last_Name"  value="<?php print "$Last_Name"; ?>" />
    </td>
  </tr>
  <tr>
    <td class="right">Email:</td>
    <td class="left left_nowrap"><input type="text" class="left left_nowrap tb10" id="Email" value="<?php print "$Email";?>" /></td>
  </tr>
  <tr>
    <td class="right">Password:</td>
    <td class="left left_nowrap"><input type="password" class="left left_nowrap tb10" id="Password"/></td>
  </tr>
  <tr>
    <td class="right">Confirm Password:</td>
    <td class="left left_nowrap"><input type="password" class="left left_nowrap tb10" id="Confirm_Password"/></td>
  </tr>
  <tr>
    <td class="right">Gender:</td>
    <td class="left left_nowrap"><span class="right">
      <select name="Gender" class="large tb10" id="Gender" value="<?php print "$gender"; ?>">
        <option value="Please Select...">Please Select...</option>
        <option value="Male">Male</option>
        <option value="Female">Female</option>
      </select>
    </span></td>
  </tr>
  <tr>
    <td class="right">&nbsp;</td>
    <td class="left"><input type="submit" class="submit tb10" value="Sign-UP" /></td>
  </tr>     
    </form>

connect_to_mysql.php

 <?php 

/* 
1: "die()" will exit the script and show an error statement if something goes wrong with the "connect" or "select" functions. 
2: A "mysql_connect()" error usually means your username/password are wrong 
3: A "mysql_select_db()" error usually means the database does not exist.
*/
// Place db host name. Sometimes "localhost" but 
// sometimes looks like this: >>      ???mysql??.someserver.net
$db_host = "localhost";
// Place the username for the MySQL database here
$db_username = "Shayaa"; 
// Place the password for the MySQL database here
$db_pass = "nestle324"; 
// Place the name for the MySQL database here
$db_name = "social_media";

// Run the actual connection here 
$con = mysql_connect("$db_host","$db_username","$db_pass") or die ("could not connect to mysql");
mysql_select_db('social_media', $con ) or die ("no database");         
?>

Answer 1

MySQL服務器可以處理許多數據庫。 至少你將擁有mysql數據庫本身，它擁有服務器特定的數據，如用戶帳戶和你自己的應用程序數據庫。

因此，僅僅連接到服務器對於與數據庫相關的任何查詢都是不夠的。 您必須選擇一個數據庫。

在您的情況下，在不修改其余代碼和MySQL API之前，您需要先使用此命令，然后才能提交查詢

mysql_select_db( 'your database name', $dbConnection );

http://php.net/manual/en/function.mysql-select-db.php

建議不要使用PHP的MySQL擴展，因為將來的PHP版本將刪除它。 請參閱上面的鏈接了解替代方案

Answer 2

• 使用mysqli或PDO

• 看看你是否在沒有選擇數據庫時選擇數據庫$sql = mysql_select_db('your database name', $connection)

• stristr達到完全相同的結果為eregi_replace （至少，當你不使用正則表達式）：

編輯：

         $sql= "INSERT INTO members (firstname, lastname, email, password,bio_body, sign_up_date , account_type) 
 VALUES(  '".$_SESSION['firstname']."','".$_SESSION['lastname']."','".$_SESSION['email']."','".$_SESSION['password']."','NULL' ,NOW() , 'a' ) " ; 

 if (!mysql_query($sql))
{
 die('Error: ' . mysql_error());
}
echo "1 record added";