回声问题与while/if/else [英] Echo issue with while / if / else

问题描述

我是学习PHP和MySQL的初学者，并且已经进入 Head First PHP&的第5章. MySQL ，并且正在尝试一个自制项目，在该项目中创建了一个数据库和一个index.php页面，从中可以看到打印出来的结果.当我进入index.php页面时，我看到了HTML标题，但是PHP代码未打印出我的提交内容.我必须假设我的代码语法正确，否则我将得到空白页.有人可以告诉我我写错了什么，但没有输出结束吗?

I am a beginner learning PHP and MySQL and have gotten to chapter 5 of Head First PHP & MySQL and am attempting a self made project in which I created a database and a index.php page where I can see the results printed out. When I go to my index.php page I see the HTML title but the PHP code is not printing out my submissions. I have to assume my code syntax is correct or I would end up with a blank page. Can someone please tell me what I have coded wrong to wind up with no output?

<?php

$dbc = mysqli_connect(localhost, root, root, itmyfamily);
$query = "SELECT * FROM itsmyfamily ORDER BY last_name ASC, first_name DESC, date ASC";
$data = mysqli_query($dbc, $query);
$i = 0;

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

推荐答案

在源代码上方设置此项以显示错误.

Set this on top of the source code to display errors.

<?php error_reporting( E_ALL ); ?>

或按如下所示在php.ini中设置display_erros:

Or set display_erros in php.ini as follows:

display_errors = On

error_reporting = E_ALL | E_STRICT

也尝试将您的源代码替换为以下内容，

Also try to replace your source code with following,

<?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);

  ?>

如果您可以这样写，我想您甚至不需要$i.

If you can write it this way, I think you don't 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的更多信息mysql_fetch_array http://www.php.net/mysql_fetch_array

Read here for more information about PHP mysql_fetch_array http://www.php.net/mysql_fetch_array

在此处阅读有关迭代的更多信息. http://webcheatsheet.com/php/loops.php

Read here for more information about iterations. http://webcheatsheet.com/php/loops.php

请注意，PHP 5.5.0不推荐使用您使用的方法.因此，我建议您考虑使用 mysqli 或 PDO .可以在下面的PHP手册链接中找到示例

Please note that the method you have used is deprecated from PHP 5.5.0. So I suggest you consider mysqli or PDO. Examples can be found in below PHP manual links

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

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

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