2015-11-08 108 views
1

我是PHP和MySql的新手。我正在尝试连接到MySql数据库并在屏幕上显示其内容。但没有显示出来。虽然我能够获得“祝贺”的信息,但我无法获得任何其他信息。请帮我弄清楚我的代码有什么问题。无法使用PHP获取MySql输出

<?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循环之前,请尝试使print_r($ result)检查是否有从查询中获取的结果。 – BoilingLime

+0

试试这个'if(!$ result)echo'找不到数据';' –

回答

1

您尚未定义$mydb变量的值。也做了其他几个更正。尝试执行下面的代码。

<?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