Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
529 views
in Technique[技术] by (71.8m points)

database - retrieve data from db and display it in table in php .. see this code whats wrong with it?

  $db = mysql_connect("localhost", "root", "");
  $er = mysql_select_db("ram");
  $query = "insert into names values('$name','$add1','$add2','$mail')";
  $result = mysql_query($query);
  print "<p> Person's Information Inserted </p>";
  $result = mysql_query("SELECT * FROM names");
?>

<table border="2">
   <tr>
      <th>Name</th>
      <th>Address Line 1</th>
      <th>Address Line 2 </th>
      <th>E-mail Id </th>
    </tr>
    <? 
    while ($array = mysql_fetch_row($result));
    {
        print "<tr> <td>";
        echo $array[0]; 
        print "</td> <td>";
        echo $array[1]; 
        print "</td> <td>";
        echo $array[2]; 
        print "</td> <td>";
        echo $array[3]; 
        print "</td> </tr>";
    }
?>
See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Try this:

<?php

 # Init the MySQL Connection
  if( !( $db = mysql_connect( 'localhost' , 'root' , '' ) ) )
    die( 'Failed to connect to MySQL Database Server - #'.mysql_errno().': '.mysql_error();
  if( !mysql_select_db( 'ram' ) )
    die( 'Connected to Server, but Failed to Connect to Database - #'.mysql_errno().': '.mysql_error();

 # Prepare the INSERT Query
  $insertTPL = 'INSERT INTO `name` VALUES( "%s" , "%s" , "%s" , "%s" )';
  $insertSQL = sprintf( $insertTPL ,
                 mysql_real_escape_string( $name ) ,
                 mysql_real_escape_string( $add1 ) ,
                 mysql_real_escape_string( $add2 ) ,
                 mysql_real_escape_string( $mail ) );
 # Execute the INSERT Query
  if( !( $insertRes = mysql_query( $insertSQL ) ) ){
    echo '<p>Insert of Row into Database Failed - #'.mysql_errno().': '.mysql_error().'</p>';
  }else{
    echo '<p>Person's Information Inserted</p>'
  }

 # Prepare the SELECT Query
  $selectSQL = 'SELECT * FROM `names`';
 # Execute the SELECT Query
  if( !( $selectRes = mysql_query( $selectSQL ) ) ){
    echo 'Retrieval of data from Database Failed - #'.mysql_errno().': '.mysql_error();
  }else{
    ?>
<table border="2">
  <thead>
    <tr>
      <th>Name</th>
      <th>Address Line 1</th>
      <th>Address Line 2</th>
      <th>Email Id</th>
    </tr>
  </thead>
  <tbody>
    <?php
      if( mysql_num_rows( $selectRes )==0 ){
        echo '<tr><td colspan="4">No Rows Returned</td></tr>';
      }else{
        while( $row = mysql_fetch_assoc( $selectRes ) ){
          echo "<tr><td>{$row['name']}</td><td>{$row['addr1']}</td><td>{$row['addr2']}</td><td>{$row['mail']}</td></tr>
";
        }
      }
    ?>
  </tbody>
</table>
    <?php
  }

?>

Notes, Cautions and Caveats

Your initial solution did not show any obvious santisation of the values before passing them into the Database. This is how SQL Injection attacks (or even un-intentional errors being passed through SQL) occur. Don't do it!

Your database does not seem to have a Primary Key. Whilst these are not, technically, necessary in all usage, they are a good practice, and make for a much more reliable way of referring to a specific row in a table, whether for adding related tables, or for making changes within that table.

You need to check every action, at every stage, for errors. Most PHP functions are nice enough to have a response they will return under an error condition. It is your job to check for those conditions as you go - never assume that PHP will do what you expect, how you expect, and in the order you expect. This is how accident happen...

My provided code above contains alot of points where, if an error has occured, a message will be returned. Try it, see if any error messages are reported, look at the Error Message, and, if applicable, the Error Code returned and do some research.

Good luck.


与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...