insert data into a table using html form

Question

I have a html form on the main page to insert data into a mysql table.

<form action ="http://127.0.0.1/insertdata.php" name="salesRecords" method="post">
Client name = <input type="text" name="client" value="" /> </br>
Date of sale = <input type="text" name="date" value="" /> (YYYY-MM-DD) </br>
Value of sale = <input type="text" name="amount" value="" /> (DDDD.DD) </br>
<input type="submit" value="Submit" onclick="return validateForm();"/>

I use the following code in receiveform.php file to display the the table data and it works correctly.

<?php

//create connection
$con = mysql_connect("localhost", "admin", "password")
//query database and write out results
$sql = 'select * from sales';
$result = mysql_query($sql, $con);
echo "<table>";
while($row = mysql_fetch_array($result))
{
    echo "<tr><td>"
    echo $row['client'];
    echo "</td><td>";
    echo $row['date'];
    echo "</td><td>";
    echo $row['amount'];
    echo "</td></tr>";
}
echo "</table>";
?>

I use the following code in an insertdata.php file. However, it doesn't work properly. Can anyone help?

/Form values
$_POST['client'];
$_POST['date'];
$_POST['amount'];

//insert data into table
$insertData = 'insert into sales (client, date, amount) values
    ('$_POST[client]', '$_POST[date]', '$_POST[amount]')';

$result = mysql_query($insertData, $con);
if($result)
{
    echo 'Data inserted successfully';
}else{
    echo 'Data insertion failed: ' .mysql_error();
}
?>

Answer 1

赞，像这样修改您的查询

$insertData = 'insert into sales (client, date, amount) values ("'.$_POST['client'].'", "'.$_POST['date'].'", "'.$_POST['amount'].'")';

Answer 2

The problem is with how you are have the $_POST array key, they should have quotes around them but this can cause problems with inputting into mysql in this way. Ram Sharma's answer will work but would suggest using PDO instead for working with databases.

Answer 3

$con = mysql_connect("localhost", "admin", "password");

After this line add

$selected = mysql_select_db("db",$con) ;

Thanks