jose Posted April 28, 2010 Share Posted April 28, 2010 Hi there, Anybody can tell why it is not possible to insert any data in MySql database? It is located in my machine, apparently everything is fine, but something is wrong and I cannot localize it. Here the script: <?php require_once('connectvars.php'); // Connect to the database $dbc = mysqli_connect('localhost', 'root', 'zam', 'entrepreneurial') or die ('Error connecting to MySQL server'); if (isset($_POST['submit'])) { // Grab the profile data from the POST $username = mysqli_real_escape_string($dbc, trim($_POST['username'])); $password1 = mysqli_real_escape_string($dbc, trim($_POST['password1'])); $password2 = mysqli_real_escape_string($dbc, trim($_POST['password2'])); if (!empty($username) && !empty($password1) && !empty($password2) && ($password1 == $password2)) { // Make sure someone isn't already registered using this username $query = "SELECT * FROM entrepreneurial_user WHERE username = '$username'"; $result = mysqli_query($dbc, $query); if ($result->num_rows == 0) { // The username is unique, so insert the data into the database $query = "INSERT INTO entrepreneurial_user (username, password1) VALUES ('username', SHA ('password1'))"; $row_cnt = $result->num; // Confirm success with the user echo '<p>Your new account has been successfully created. You\'re now ready to <a href="login.php">log in</a>.</p>'; mysqli_close($dbc); exit(); } else { // An account already exists for this username, so display an error message echo '<p class="error">An account already exists for this username.</p>'; $username = ""; } } else { echo '<p class="error">You must enter all of the sign-up data, including the desired password twice.</p>'; } } mysqli_close($dbc); ?> Thanks Quote Link to comment Share on other sites More sharing options...
Hybride Posted April 28, 2010 Share Posted April 28, 2010 Please use the proper code tags next time when you post code, thanks. Is that all of the code? What about the actual form? Quote Link to comment Share on other sites More sharing options...
jose Posted April 28, 2010 Author Share Posted April 28, 2010 the rest of the code <!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" xml:lang="en" lang="en"> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8" /> <title> Entrepre - Sign Up</title> <link rel="stylesheet" type="text/css" href="style.css" /> <?php require_once('connectvars.php'); // Connect to the database $dbc = mysqli_connect('localhost', 'root', 'zamb', 'entrepreneuria') or die ('Error connecting to MySQL server'); if (isset($_POST['submit'])) { // Grab the profile data from the POST $username = mysqli_real_escape_string($dbc, trim($_POST['username'])); $password1 = mysqli_real_escape_string($dbc, trim($_POST['password1'])); $password2 = mysqli_real_escape_string($dbc, trim($_POST['password2'])); if (!empty($username) && !empty($password1) && !empty($password2) && ($password1 == $password2)) { // Make sure someone isn't already registered using this username $query = "SELECT * FROM entrepreneurial_user WHERE username = '$username'"; $result = mysqli_query($dbc, $query); if ($result->num_rows == 0) { // The username is unique, so insert the data into the database $query = "INSERT INTO entrepreneurial_user (username, password1) VALUES ('username', SHA ('password1'))"; $row_cnt = $result->num; // Confirm success with the user echo '<p>Your new account has been successfully created. You\'re now ready to <a href="login.php">log in</a>.</p>'; mysqli_close($dbc); exit(); } else { // An account already exists for this username, so display an error message echo '<p class="error">An account already exists for this username.</p>'; $username = ""; } } else { echo '<p class="error">You must enter all of the sign-up data, including the desired password twice.</p>'; } } mysqli_close($dbc); ?> <style type="text/css"> <!-- body,td,th { font-size: 16px; font-family: Tahoma, Geneva, sans-serif; } #wrapper { width: 860px; margin-right: auto; margin-left: auto; padding: 10px; background-color: #FFF; font-size: 12px; } body,td,th { font-family: Tahoma, Geneva, sans-serif; color: #000; background-color: #FFF; } body { background-color: #E6FFFE; .regist { font-weight: bold; } .regist { font-weight: bold; font-size: 18pt; } .u { font-size: 12pt; } .user { font-size: 12px; } .user { font-size: 12pt; } .signup { font-size: 12pt; font-weight: bold; } </style></head> <body> <div id="wrapper"> <p><img src="images/entrepreneurs banner.jpg" width="860" height="60" alt="banner" /></p> <?php // If the cookie is empty, show any error message and the log-in form; otherwise confirm the log-in if (empty($_COOKIE['user_id'])) { echo '<p class="error">' . $error_msg . '</p>'; ?> <form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset> <legend class="sign">Sign Up </legend> <table width="400" border="0"> <tr> <td class="user" style="font-size: 12pt"><label for="username">Username:</label></td> <td><input name="username" type="text" value="<?php if (!empty($user_username)) echo $user_username; ?>" size="25" maxlength="15" /> <br /></td> </tr> <tr> <td class="user" style="font-size: 12pt"><label for="password1">Password:</label></td> <td><input type="password" id="password1" name="password1" /> <br /></td> </tr> <tr> <td class="user" style="font-size: 12pt"><label for="password2">Password (retype):</label></td> <td><input type="password" id="password2" name="password2" /> <br /></td> </tr> <tr> <td> </td> <td><input type="submit" value="Sign Up" name="submit" /></td> </tr> </table> </fieldset> </form> <table width="662" border="0"> <tr> <?php } else { // Confirm the successful log-in echo('<p class="login">You are logged in as ' . $_COOKIE['username'] . '.</p>'); } ?> </body> </html> Quote Link to comment Share on other sites More sharing options...
jose Posted April 28, 2010 Author Share Posted April 28, 2010 Hi there, Anybody can tell why it is not possible to insert any data in MySql database? It is located in my machine, apparently everything is fine, but something is wrong and I cannot localize it. Here the script: <?php require_once('connectvars.php'); // Connect to the database $dbc = mysqli_connect('localhost', 'root', 'zam', 'entrepreneurial') or die ('Error connecting to MySQL server'); if (isset($_POST['submit'])) { // Grab the profile data from the POST $username = mysqli_real_escape_string($dbc, trim($_POST['username'])); $password1 = mysqli_real_escape_string($dbc, trim($_POST['password1'])); $password2 = mysqli_real_escape_string($dbc, trim($_POST['password2'])); if (!empty($username) && !empty($password1) && !empty($password2) && ($password1 == $password2)) { // Make sure someone isn't already registered using this username $query = "SELECT * FROM entrepreneurial_user WHERE username = '$username'"; $result = mysqli_query($dbc, $query); if ($result->num_rows == 0) { // The username is unique, so insert the data into the database $query = "INSERT INTO entrepreneurial_user (username, password1) VALUES ('username', SHA ('password1'))"; $row_cnt = $result->num; // Confirm success with the user echo '<p>Your new account has been successfully created. You\'re now ready to <a href="login.php">log in</a>.</p>'; mysqli_close($dbc); exit(); } else { // An account already exists for this username, so display an error message echo '<p class="error">An account already exists for this username.</p>'; $username = ""; } } else { echo '<p class="error">You must enter all of the sign-up data, including the desired password twice.</p>'; } } mysqli_close($dbc); ?> Thanks<form method="post" action="<?php echo $_SERVER['PHP_SELF']; ?>"> <fieldset> <legend class="sign">Sign Up </legend> <table width="400" border="0"> <tr> <td class="user" style="font-size: 12pt"><label for="username">Username:</label></td> <td><input name="username" type="text" value="<?php if (!empty($user_username)) echo $user_username; ?>" size="25" maxlength="15" /> <br /></td> </tr> <tr> <td class="user" style="font-size: 12pt"><label for="password1">Password:</label></td> <td><input type="password" id="password1" name="password1" /> <br /></td> </tr> <tr> <td class="user" style="font-size: 12pt"><label for="password2">Password (retype):</label></td> <td><input type="password" id="password2" name="password2" /> <br /></td> </tr> <tr> <td> </td> <td><input type="submit" value="Sign Up" name="submit" /></td> </tr> </table> </fieldset> </form> <table width="662" border="0"> <tr> Thanks Quote Link to comment Share on other sites More sharing options...
Deoctor Posted April 28, 2010 Share Posted April 28, 2010 i hope u r missing the basic thing of using the mysql_query in this one $query = "INSERT INTO entrepreneurial_user (username, password1) VALUES ('username', SHA ('password1'))"; $row_cnt = $result->num; Quote Link to comment Share on other sites More sharing options...
jose Posted April 29, 2010 Author Share Posted April 29, 2010 i hope u r missing the basic thing of using the mysql_query in this one $query = "INSERT INTO entrepreneurial_user (username, password1) VALUES ('username', SHA ('password1'))"; $row_cnt = $result->num; What do you mean? Quote Link to comment Share on other sites More sharing options...
Deoctor Posted April 30, 2010 Share Posted April 30, 2010 what i mean is that $query = "INSERT INTO entrepreneurial_user (username, password1) VALUES ('username', SHA ('password1'))"; this code should be accompanied with the $result = mysqli_query($dbc, $query); Quote Link to comment Share on other sites More sharing options...
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