Bir tablo oluşturmak için komut dosyası ve SQL alışkanlık iş alanlar

1 Cevap php

Bu Uyarı lenghty olduğunu! Eğer knowledagble eğer saldırı. iyi en azından benim gibi bir acemi newb daha sonra.

Bu script aşağıda ayrıntılı olarak üç dosya kullanır. Bu form girişi gelen veritabanı ve alanları oluşturmak suppoed edilir. Bu sonuna kadar alır ve my_contacts oluşturuldu gösterir!. PhpMyAdmin gitmek Ama tablo yaratılmış değil.

Ben MySQL bir tablo oluşturmak için kullanılan bir dosya adında show_createtable.html var

<!DOCTYPE html PUBLIC "-//W3C//DTD XHTML 1.0 Transitional//EN" "http://www.w3.org/TR/xhtml1/DTD/xhtml1-transitional.dtd">
<html xmlns="http://www.w3.org/1999/xhtml">
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Untitled Document</title>
</head>

<body>
<h1>Step 1: Name and Number</h1>
<form method="post" action="do_showfielddef.php" />
<p><strong>Table Name:</strong><br />
<input type="text" name="table_name" size="30" /></p>
<p><strong>Number of fields:</strong><br />
<input type="text" name="num_fields" size="30" /></p>
<p><input type="submit" name="submit" value="go to step2" /></p>
</form>


</body>
</html>

Bu Form Mesajlar do_showfielddef.php için

    <?php
    //validate important input
    if ((!$_POST[table_name]) || (!$_POST[num_fields])) {
        header( "location: show_createtable.html");
               exit;
    }

    //begin creating form for display
    $form_block = "
    <form action=\"do_createtable.php\" method=\"post\">
    <input name=\"table_name\" type=\"hidden\" value=\"$_POST[table_name]\">
    <table cellspacing=\"5\" cellpadding=\"5\">
      <tr>
        <th>Field Name</th><th>Field Type</th><th>Table Length</th><th>Primary Key?</th><th>Auto-Increment?</th>
      </tr>";

    //count from 0 until you reach the number fo fields
    for ($i = 0; $i <$_POST[num_fields]; $i++) {
      $form_block .="
      <tr>
      <td align=center><input type=\"texr\" name=\"field name[]\"
      size=\"30\"></td>
      <td align=center>
        <select name=\"field_type[]\">
            <option value=\"char\">char</option>
            <option value=\"date\">date</option>
            <option value=\"float\">float</option>
            <option value=\"int\">int</option>
            <option value=\"text\">text</option>
            <option value=\"varchar\">varchar</option>
            </select>
      </td>
      <td align=center><input type=\"text\" name=\"field_length[]\" size=\"5\"></td>
      <td aligh=center><input type=\"checkbox\" name=\"primary[]\" value=\"Y\"></td>
      <td aligh=center><input type=\"checkbox\" name=\"auto_increment[]\" value=\"Y\"></td>

    </tr>";
    }

    //finish up the form 
    $form_block .= "
    <tr>
        <td align=center colspan=3><input type =\"submit\" value=\"create table\">
        </td>
    </tr>
    </table>
    </form>";

    ?>

    <html>
    <head>
    <meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
    <title>Create a database table: Step 2</title>
    </head>

    <body>
    <h1>defnie fields for <? echo "$_POST[table_name]"; ?> 
    </h1>
    <? echo "$form_block"; ?>

    </body>
    </html>

Which in turn creates the table and fields with this file do_showfielddef.php 

//connect to database
$connection = @mysql_connect("localhost", "user", "pass")
    or die(mysql_error());
$db = @mysql_select_db($db_name, $connection)
    or die(mysql_error());

//start creating the SQL statement
$sql = "CREATE TABLE $_POST[table_name](";
    //continue the SQL statement for each new field                                 
    for ($i = 0; $i < count($_POST[field_name]); $i++) {
        $sql .= $_POST[field_name][$i]." ".$_POST[field_type][$i];

    if ($_POST[auto_increment][$i] =="Y") {
        $additional = "NOT NULL auto_increment";
    } else {
        $additional = "";
    }

    if ($_POST[primary][$i] =="Y") {
        $additional .= ", primary key (".$_POST[field_name][$i].")";
    } else {
        $additional = "";
    }

    if ($_POST[field_length][$i] !="") {
        $sql .= " (".$_POST[field_length][$i].") $additional ,";
        } else {
            $sql .=" $additional ,";
        }
    }

//clean up the end of the string
$sql = substr($sql, 0, -1);
$sql .= ")";

//execute the query
$result = mysql_query($sql, $connection) or die(mysql_error());
//get a giid message for display upon success
if ($result) {
    $msg = "<p>" .$_POST[table_name]." has been created!</p>";
}
?>

<html>
<head>
<meta http-equiv="Content-Type" content="text/html; charset=utf-8" />
<title>Create A Database Table: Step 3</title>
</head>

<body>
<h1>Adding table to <? echo "$db_name"; ?>...</h1>
<? echo "$msg"; ?>
</body>
</html>

1 Cevap

I cant believe I went to all the trouble of wrinting this Question. I had another good look at the phpMYAdmin and it had worked. The table had been created under a database called testDB which I assumed had nothing it in. How did the script decided to etner this as a child under the testDB database?

Lütfen giriş için bir kez daha teşekkürler herkese, Bu site gerçekten amazine ve benim öz gibi bir acemi için çok değerlidir.

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