2015-11-08 91 views
1

我是PHP和MySql的新手。我正在嘗試連接到MySql數據庫並在屏幕上顯示其內容。但沒有顯示出來。雖然我能夠獲得「祝賀」的信息,但我無法獲得任何其他信息。請幫我弄清楚我的代碼有什麼問題。無法使用PHP獲取MySql輸出

<?php 
    $dbhost= "localhost"; 
    $dbuser= "root"; 
    $dbpass= ""; 
    $dbconnect= mysqli_connect('$dbhost', '$dbuser', '$dbpass','$mydb'); 
    /*mysqli_select_db($mydb);*/ 
     if(!mysqli_select_db){ 
      echo "database not found"; 
      die(mysqli_error()); 
     } 
     else{ 
      echo "congrats"; 
     } 


    $query = "SELECT * FROM userinfo"; 
    $result = mysqli_query($dbconnect, $query); 

    while($record = mysqli_fetch_array($result)){ 
     echo $record['Name'], $record['Email'], $record['Contact']; 
    } 

?> 
+0

在您的while循環之前,請嘗試使print_r($ result)檢查是否有從查詢中獲取的結果。 – BoilingLime

+0

試試這個'if(!$ result)echo'找不到數據';' –

回答

1

您尚未定義$mydb變量的值。也做了其他幾個更正。嘗試執行下面的代碼。

<?php 
    $dbhost= "localhost"; 
    $dbuser= "root"; 
    $dbpass= ""; 
    $mydb = "your_dbname"; 
    $dbconnect= mysqli_connect($dbhost, $dbuser, $dbpass,$mydb); 
    /*mysqli_select_db($mydb);*/ 
     if(!mysqli_select_db){ 
      echo "database not found"; 
      die(mysqli_error()); 
     } 
     else{ 
      echo "congrats"; 
     } 


    $query = "SELECT * FROM userinfo"; 
    $result = mysqli_query($dbconnect, $query); 

    while($record = mysqli_fetch_array($result)){ 
     echo $record['Name']." ".$record['Email']." ".$record['Contact']; 
    } 

?> 
+0

thanx sandeepsure,它工作... – Sachin