I have a problem with this code. It has syntax error and I don't know what is it.
<?php
$host = 'localhost';
$user = 'root';
$pass = '';
$db = 'id1381007_accounts';
$conn = new mysqli($host,$user,$pass,$db) or die($mysqli->error);
if (!$conn) {
die('Could not connect: ' . mysql_error());
$sql = 'SELECT id FROM users WHERE email=\"donat12#icloud.com\"';
echo $sql;
?>
There are some issue with the code. First you forgot to close the if condition over here
if (!$conn) {
And then you forgot to execute the sql query
the complete code would be like
<?php
$host = 'localhost';
$user = 'root';
$pass = '';
$db = 'id1381007_accounts';
$conn = new mysqli($host,$user,$pass,$db) or die($mysqli->error);
if (!$conn) {
die('Could not connect: ' . mysql_error());
}
$sql = 'SELECT id FROM users WHERE email=\"donat12#icloud.com\"';
if ($result = $conn->query($sql)) {
while ( $row = $result->fetch_assoc()) {
$data[] = $row;
}
echo "<pre>";
print_r($data);
echo "</pre>";
}
$conn->close();
?>
There are two errors
You are missing } closing bracket after die
Mysql query is wrong.
So the code should be
if (!$conn) {
die('Could not connect: ' . mysql_error());
}
$sql = 'SELECT id FROM users WHERE email="donat12#icloud.com"';
echo $sql;
Related
I have written following code to connect mysql in php but I am not getting output.
<?php
$servername = "localhost";
$username = "root";
$password = "pravin";
$mysql_conn = new mysqli($servername, $username, $password);
if ($mysql_conn->connect_error) {
die("Connection failed: ". $mysql_conn->connect_error);
}
echo "Connected successfully";
$name = $_POST["microorganism"];
echo $name;
$db_selected = mysql_select_db('yieldofvanillin', $mysql_conn);
if (!$db_selected){
die ('Can\'t use : ' . mysql_error());
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
$result = $mysql_query($query);
while ($line = myql_fetch_array($result, MYSQL_ASSOC)) {
echo $line["Substrate"];
echo $line["products"];
echo $line["Microorganism"];
echo $line["yield"];
echo $line["Reference"];
}
mysql_close($mysql_conn);
?>
The database name is "yieldofvanillin" and it has five column. I an getting output Connected successfully. After that no output. Please let me know the bug in code.
i have remove errors. which i mention in comments. Code Reference PHP Manual. you should read this manual (strongly recommended)
<?php
$mysqli = new mysqli("localhost", "root", "pravin", "yieldofvanillin");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
if ($result = $mysqli->query($query)) {
/* fetch associative array */
while ($row = $result->fetch_assoc()) {
echo $row["Substrate"];
echo $row["products"];
echo $row["Microorganism"];
echo $row["yield"];
echo $row["Reference"];
}
/* free result set */
$result->free();
}
you're mixing mysqli and mysql libraries.
the code should be:
<?php
$servername = "localhost";
$username = "root";
$password = "pravin";
$mysql_conn = new mysqli($servername, $username, $password);
if (mysqli_connect_errno()) {
die("Connection failed: ". mysqli_connect_error());
}
echo "Connected successfully";
$name = $_POST["microorganism"];
echo $name;
$db_selected = mysqli_select_db($mysql_conn,'yieldofvanillin');
if (!$db_selected){
die ('Can\'t use : ' . mysqli_error($mysql_conn));
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
$result = mysqli_query( $mysql_conn,$query);
while ($line = mysqli_fetch_assoc($result)) {
echo $line["Substrate"];
echo $line["products"];
echo $line["Microorganism"];
echo $line["yield"];
echo $line["Reference"];
}
mysqli_close($mysql_conn);
?>
Remove error in your code.Read carefully php manual.
<?php
$servername = "localhost";
$username = "root";
$password = "pravin";
$db = "yieldofvanillin";
// Create connection
$mysqli = new mysqli($servername, $username, $password, $db);
/* connection string*/
if ($mysqli->connect_errno) {
die("Connection failed: " . $mysqli->connect_error);
exit();
}
$query = "SELECT * FROM vanillin WHERE Microorganism = '$name' ";
if ($result = $mysqli->query($query)) {
while ($row = $result->fetch_assoc()) {
echo $row["Substrate"];
}
$result->free();
}
$mysqli->close();
?>
Your output not showing because mysql_fetch_array is not correct.Because you are mixing mysql_ and mysqli_ functions and you called myql_fetch_array that doesn't exist in mysqli. MySQL and MySQLi are two different PHP extensions and they cannot be mixed. Because the former is deprecated in mysqli
<?php
//connecting to the database
define('DB_HOST', 'localhost');
define('DB_NAME', 'visitor_list');
define('DB_USER','root');
define('DB_PASSWORD','');
$con=mysql_connect(DB_HOST,DB_USER,DB_PASSWORD) or die("Failed to connect to MySQL: " . mysql_error()); $db=mysql_select_db(DB_NAME,$con) or die("Failed to connect to MySQL: " .
mysql_error()); //inserting Record to the database
$dat1 = date("d");
$moth1 = date("m");
$year1 = date("Y");
$sql2 = 'SELECT * FROM list1 ORDER BY card_no DESC LIMIT 1;';
$retval = mysql_query($con,$sql2 );
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
/*echo "{$row['card_no']}";*/
$in= $row['card_no'];
$in++;
}
$ap_ty = $_POST['ap_ty'];
$per = $_POST['per'];
$no_vis = $_POST['no_vis'];
$nm_vis = $_POST['nm_vis'];
$co_nm = $_POST['co_nm'];
$ad = $_POST['ad'];
$po = $_POST['po'];
$ty_vis = $_POST['ty_vis'];
$met_nm = $_POST['met_nm'];
$dep = $_POST['dep'];
$des = $_POST['des'];
$por_met = $_POST['por_met'];
$vad = $_POST['vad'];
if(!isset($_POST['submit'])){
$query = "INSERT INTO list1 (card_no,d1,m1,y1,app_ty,no_per,area_vis,nm,com_nm,add1,pho,ty1_vis,met_with,depart,desi,pur_meet,vad_up) VALUES ('$in','$dat1','$moth1','$year1','$ap_ty','$per','$no_vis','$nm_vis','$co_nm','$ad','$po','$ty_vis','$met_nm','$dep','$des','$por_met','$vad')";}
$result = mysql_query($query);
if($result)
{
echo '<script type="text/javascript">
alert("Your Deatils has been Submitted");
</script> ';
header('location:index.php');
}
else
{ die('Error: '.mysql_error($con)); }
mysql_close($con);
?>
I want to save the $in variables value in the query but not able to save it. At time of insertion it saves the value 0 not the actual value.
<input type="text" class="form-control" name="vis_id" value=" <?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$sql = 'SELECT * FROM list1 ORDER BY card_no DESC LIMIT 1;';
mysql_select_db('visitor_list');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
/*echo "{$row['card_no']}";*/
$in= $row['card_no'];
$in+
$test = $in;
}
mysql_close($conn);
?>" disabled>
this is the actual code
in this input box the value is incrementing
help me save the value of this input box into another file which is database file.
$query = "SELECT * FROM Student WHERE student_id = $child AND school_id = $college";
$result = mysqli_query($conic, $query);
if(!$result){
die(" query failed");
echo 'error here';
} else {
i am trying to retrieve some information from the SQL database but the query seems to fail any ideas?
You must learn more : how to retrieve data from database using select in php
so try out following example to display all the records from Student table
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = 'rootpassword';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$sql = "SELECT * FROM student WHERE student_id = '"$child"' AND school_id = '"$college"' ";
mysql_select_db('test_db');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
// some code here
}
echo "Fetched data successfully\n";
mysql_close($conn);
?>
NOTE
The mysql extension is deprecated and will be removed in the future:
use mysqli or PDO
<?php
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = "";
$name=$_POST['user'];
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$sql = 'SELECT * FROM `usertable` WHERE `username`='$name'' ;
mysql_select_db('myxiv');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not get data: ' . mysql_error());
}
while($row = mysql_fetch_array($retval, MYSQL_ASSOC))
{
echo "Name :{$row['username']} .<br> ";
}
mysql_close($conn);
?>
Here is my code and i'm getting this error.
Syntax error, unexpected '$name' (T_VARIABLE)
I m trying to retrive data from a table to display in the profile page
$sql = 'SELECT * FROM usertable WHERE username='$name'' ;
Replace Your above query by bellow code and check it.
$sql = "SELECT * FROM usertable WHERE username=' ".$name." ' ";
Try this
Change
$sql = 'SELECT * FROM usertable WHERE username`='$name'' ;
TO
$sql = "SELECT * FROM usertable WHERE username='".$name."';
You make a few mistakes. mysql (which is depracted (take a look at PDO or mysqli)) doesn't need a connection.
You could have a double single quote error but the following should fix some of it:
if($_POST) {
$dbhost = 'localhost';
$dbuser = 'root';
$dbpass = '';
$name = mysql_real_escape_string(strip_tags($_POST['user']));
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn ) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db('myxiv');
$retval = mysql_query("SELECT * FROM `usertable` WHERE `username` = '$name'");
if(! $retval ) {
die('Could not get data: ' . mysql_error());
} else {
while($row = mysql_fetch_array($retval, MYSQL_ASSOC)) {
echo "Name :{$row['username']} .<br> ";
}
}
}
im getting the following syntax error can someone please help!
im guessing it something soooo easy but i have been looking at it for ages and can see what im doing wrong
<?php
if(isset($_POST['delete']))
{
$dbhost = '';
$dbuser = '';
$dbpass = '';
$conn = mysql_connect($dbhost, $dbuser, $dbpass);
if(! $conn )
{
die('Could not connect: ' . mysql_error());
}
$CourseId = $_POST['CourseId'];
$sql = "DELETE course ".
" WHERE CourseId = $CourseId" ;
mysql_select_db('d11os_projectdb');
$retval = mysql_query( $sql, $conn );
if(! $retval )
{
die('Could not delete data: ' . mysql_error());
}
echo "Deleted data successfully\n";
mysql_close($conn);
}
else
{
?>`enter code here`
$sql = "DELETE FROM course ". --<-- Missing key word FROM
" WHERE CourseId = $CourseId"
You are missing the table name from the sql query
$sql = "DELETE course FROM **table_name**".
" WHERE CourseId = $CourseId" ;