Question

我是初学者学习php和mysql并且已经获得了Head First PHP＆amp; amp; MySQL和我正在尝试一个自制项目，我在其中创建了一个数据库和一个index.php页面，在那里我可以看到打印出的结果。当我转到我的index.php页面时，我看到了HTML标题但是php代码没有打印出我提交的内容。我必须假设我的代码语法是正确的，否则我最终会得到一个空白页面。有人可以告诉我，我编码错误的结果没有输出吗？感谢。

<?php

//Establish connection with database

$dbc = mysqli_connect(localhost,root,root,itmyfamily);

//Order the data to be retrieved

$query = "SELECT * FROM itsmyfamily ORDER BY last_name ASC, first_name `enter code here`DESC, date ASC";

//Execute the connect command and the query 
$data = mysqli_query($dbc,$query);
//Loop through the array of family submissions, formatting it as html
$i = 0;

while ($row = mysqli_fetch_array($data)) {

//Display family submissions

    if ($i == 0) {

        echo '<strong>First Name:</strong> ' .$row['first_name']. ' <br/>';
        echo '<strong>Last Name:</strong> ' .$row['last_name']. ' <br />';
        echo '<strong>Spouse Name:</strong> ' .$row['spouse_name']. ' <br/>';
        echo '<strong>Email:</strong> ' .$row['email']. ' <br />';

        }
        else {
            echo 'There is no info in the database';        
        }

         $i++;
    }


    mysqli_close($dbc);

   ?>

Answer 1

将此设置在源代码之上以显示错误。

<?php error_reporting( E_ALL ); ?>

或者在php.ini中设置display_erros，如下所示：

display_errors = On

error_reporting = E_ALL | E_STRICT

还尝试使用以下内容替换源代码，

<?php

  //Establish connection with database

  $dbc = mysqli_connect(localhost,root,root,itmyfamily);

  //Order the data to be retrieved

  $query = "SELECT * FROM itsmyfamily ORDER BY last_name ASC, first_name DESC, date ASC";

  //Execute the connect command and the query 

$data = mysqli_query($dbc,$query);


  while ($row = mysqli_fetch_array($data)) {


//Loop through the array of family submissions, formatting it as html

  $i = 0;

  //Display family submissions

    if ($i == 0) {

    echo '<strong>First Name:</strong> ' .$row['first_name']. ' <br />';
    echo '<strong>Last Name:</strong> ' .$row['last_name']. ' <br />';
    echo '<strong>Spouse Name:</strong> ' .$row['spouse_name']. ' <br />';
    echo '<strong>Email:</strong> ' .$row['email']. ' <br />';

}
    else{
        echo 'There is no info in the database';       
  }
  $i++;


  }


    mysqli_close($dbc);

  ?>

<强>更新

If you can write it this way, i think you dont even need the $i.

$result = mysql_query("SELECT id, first_name FROM mytable");

if($result){

while ($row = mysql_fetch_array($result, MYSQL_ASSOC)) {
    printf("ID: %s  Name: %s", $row["id"], $row["first_name"]);
}

}
else
 echo 'There is no info in the database';

在此处阅读有关php mysql_fetch_array http://www.php.net/mysql_fetch_array

的更多信息

请阅读此处以获取有关迭代的更多信息。 http://webcheatsheet.com/php/loops.php

请注意，您使用的方法已从PHP 5.5.0中弃用。所以我建议你考虑 mysqli 或 PDO 。示例可以在下面的php手册链接中找到

http://www.php.net/manual/en/mysqli.query.php

http://www.php.net/manual/en/pdo.query.php

与/ if / else的回声问题

1 个答案: