PHP从MySQL数据库错误中检索数据

Question

I have two php files that are called form1.php, and demo12.php. This is the result of the form1.php file:

 <form action="demo12.php" method="post" /> <p>Name <input type="text" name="input1" /></p> <p>Age <input type="text" name="input2" /></p> <input type="submit" value="Submit" /> </form> 

Then the code of my demo12.php is:

<?php

define('DB_NAME', 'form1');
define('DB_USER', 'root');
define('DB_PASSWORD', 'root');
define('DB_HOST', 'localhost');

$link = mysql_connect(DB_HOST, DB_USER, DB_PASSWORD);

if (!$link) {
    die('Could not connect: ' . mysql_error());
}

$db_selected = mysql_select_db(DB_NAME, $link);

if (!$db_selected) {
    die('Can\'t use ' . DB_NAME . ': ' . mysql_error());
}

$value = $_POST['input1'];
$value2 = $_POST['input2'];

$sql = "INSERT INTO user_info (name, age) VALUES ('$value', '$value2')";

if (!mysql_query($sql)) {
    die('Error: ' . mysql_error());
}

$result = mysql_query("SELECT * FROM name") or die(mysql_error());

while($row = mysql_fetch_assoc($result)) {
    echo $row["name"];
}

?>

I am able to enter a name, age, then click the submit button, which brings the user to the demo12.php page and inserts the data into the MySQL database. The problem is, I want to retrieve and then display all from the name column onto the page. So I have a database named form1, them a table named user_info. In that the 3 columns I have are ID, name, and age. To display all from the name column I have this at the near-end of my script:

$result = mysql_query("SELECT * FROM name") or die(mysql_error());

while($row = mysql_fetch_assoc($result)) {
    echo $row["name"];
}

Then when the data is inserted, then when it is going to display the all of the names in the table it has an error:

Table 'form1.name' doesn't exist

But it has no errors when I take that code out. Why is it not working?

Answer 1

You are inserting into the database user_info and then trying to select form the database name . Try changing your select to "SELECT * FROM user_info" .

Answer 2

You should have written:

$result = mysql_query("SELECT * FROM user_info") or die(mysql_error());

and not

$result = mysql_query("SELECT * FROM name") or die(mysql_error());

Answer 3

Please replace at your last part with following

$result = mysql_query("SELECT name FROM user_info") or die(mysql_error());

while($row = mysql_fetch_assoc($result))
{
    echo $row['name'];
}