When I run the code in localhost server it works properly but when we run the same code in our live server (Godaddy linux shared server) it gives sql connection failed error.
The database is quite big and in our local server the query takes approximately 2 minutes to return data, so in shared godaddy server it must take more time.
<?php
$conn = mysqli_connect("localhost","root","","doreme_eshop");
// $mysqli = new mysqli("localhost", "root", "", "doreme_eshop");
// if ($mysqli->connect_errno) {
// echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") " . $mysqli->connect_error;
// }
if(isset($_POST['submit'])){
$style=$_REQUEST['style_no'];
$result = mysqli_query($conn,"CALL style_report('".$style."')") or die("Query Failed: " . mysqli_error($conn));
//var_dump($result->fetch_assoc());
echo "<table border='1'>
<tr>
<th>total_set</th>
<th>style_no</th>
<th>size_id</th>
<th>size_description</th>
<th>Company_name</th>
<th>generate_no</th>
<th>order_Date</th>
<th>product_id</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['total_set'] . "</td>";
echo "<td>" . $row['style_no'] . "</td>";
echo "<td>" . $row['size_id'] . "</td>";
echo "<td>" . $row['size_description'] . "</td>";
echo "<td>" . $row['Company_name'] . "</td>";
echo "<td>" . $row['generate_no'] . "</td>";
echo "<td>" . $row['order_Date'] . "</td>";
echo "<td>" . $row['product_id'] . "</td>";
echo "</tr>";
}
echo "</table>";
}
?>
<html>
<head></head>
<body>
<form role="form" action="<?php echo $_SERVER['PHP_SELF']; ?>" method="POST" enctype="multipart/form-data">
<select name="style_no" >
<option value="">STYLE NO</option>
<?php
$style_result = mysqli_query($conn,"CALL get_all_style_no()") or die("Query Failed: " . mysqli_error($conn));
$row1 = mysqli_num_rows($style_result);
while ($row1 = mysqli_fetch_array($style_result)){
echo "<option value='". $row1['style_no'] ."'>" .$row1['style_no'] ."</option>" ;
}
?>
</select>
<input type="submit" value="Submit" name="submit"/>
</form>
</body>
</html>
You need to set time out on connection, secondly u can set the config file to be separated and require_once.
Try check if the DB has mysqli library,... just open connection only then close and check error
Related
I am using simple code to retrieve data from database table but not getting result it always shows "No Result found".
Table Structure
rollno Varchar(50) Primary Key,
name Varchar(100),
fname Varchar(100),
mname Varchar(100),
course Varchar(100),
duration Varchar(100),
address Varchar(100),
image blob.
HTML FORM CODE
<form name="input" action="q.php" target="display" method="post" >
Roll No: <input type="text" name="name">
<input type="submit" name="submit" value="Submit">
</form>
PHP CODE
<?php
if (isset($_POST['name'])) {
$con=mysqli_connect("mysql.1freehosting.com","u890130056_certi","samsungk2","u890130056_certi");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$name = htmlspecialchars($_POST['rollno']);
{
$result = mysqli_query($con,"SELECT * FROM certificate where rollno ='$name'");
if(mysqli_num_rows($result)>0){
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['rollno'] . "</td>";
echo "<td>" . $row['course'] . "</td>";
echo "<td>" . $row['duration'] . "</td>";
echo "<td>" . $row['name'] . "</td>";
echo "<td>" . $row['fname'] . "</td>";
echo "<td>" . $row['mname'] . "</td>";
echo "<td>" . $row['address'] . "</td>";
echo "<td>";?><img src="<?php echo $row["image"]; ?> " height="100" width="100"> <?php echo "</td>";
echo "</tr>";
}
}
else
{
echo "<tr><td colspan='4'> No Data Found , Please check your registration no. or contact the institute for clarification. ".$line.'</td></tr>';
}
mysqli_close($con);
}}
?>
name = htmlspecialchars($_POST['rollno']);
where are you getting this 'rollno??
i dont get why are you saving the result by posting['rollno'] because the name of your feild is 'name'.
change this line
$name = htmlspecialchars($_POST['rollno']);
to
$name = htmlspecialchars($_POST['name']);
I would like to know how can i put the table from mysql in a new file.php.
I want the MySql table to be on the page.
This is my code that inserts data in MySql.
<?php
// Create connection
$con = mysqli_connect("host", "id_", "password", "xxxxxx");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$Task = $_POST['Task'];
$Date = $_POST['Date'];
$Desc = $_POST['Desc'];
$sql = "INSERT INTO tasklist (Task, Date, Description)
VALUES ('$Task', '$Date', '$Desc')";
if (!mysqli_query($con, $sql)) {
die('Error: ' . mysqli_error($con));
}
echo "1 record added";
mysqli_close($con);
?>
<html>
<body>
<form action="addtask.php" method="post">
Task: <input type="text" name="Task">
Date: <input type="text" id="datepicker" name="Date">
Decrption:<textarea type="text" name="Desc"></textarea>
<input type="submit" value="submit">
</form>
</body>
</html>
Try this code:
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
also u can try w3schools sample code :
Display the Result in an HTML Table
The following example selects the same data as the example above, but will display the data in an HTML table:
<?php
$con=mysqli_connect("example.com","peter","abc123","my_db");
// Check connection
if (mysqli_connect_errno()) {
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Persons");
echo "<table border='1'>
<tr>
<th>Firstname</th>
<th>Lastname</th>
</tr>";
while($row = mysqli_fetch_array($result)) {
echo "<tr>";
echo "<td>" . $row['FirstName'] . "</td>";
echo "<td>" . $row['LastName'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
The output of the code above will be:
first code comes from "Jonnny" in this article
I'm trying to update a table given user input. Once the user hits submit on the form, I want the WHERE portion of my query to reflect the zip code entered by the user. Here is what I have so far, but it doesn't work. Any help would be greatly appreciated!
<form id="user-location" method="post" action="#">
<input id="addressInput" name="addressInput" type="text">
<input id="submit" onclick="searchLocations()" value="GO" type="button">
</form>
<?php
$con=mysqli_connect("localhost","######","######","######");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
$result = mysqli_query($con,"SELECT * FROM Prospects WHERE zip = 'echo $_POST['addressInput']'");
echo "<table width='540' cellpadding='0' border='0' cellspacing='0'>
<tr>
<th>Under 4</th>
<th>5 - 9</th>
<th>10 - 14</th>
<th>15 - 17</th>
<th>18 - 20</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['cy_pop_04'] . "</td>";
echo "<td>" . $row['cy_pop_59'] . "</td>";
echo "<td>" . $row['cy_pop_1014'] . "</td>";
echo "<td>" . $row['cy_pop_1517'] . "</td>";
echo "<td>" . $row['cy_pop_1820'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
?>
Change <input id="submit" onclick="searchLocations()" value="GO" type="button"> to <input id="submit" value="GO" type="submit" name="submit"> then use a conditional statement.
I.e.: if(isset($_POST['submit']))
Here is a prepared statement method.
The way you're doing it now (or intended to use), will leave you open to SQL injection.
<?php
$con=mysqli_connect("localhost","######","######","######");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
if(isset($_POST['submit'])){
$zip = $_POST['addressInput'];
if($query = $con->prepare("SELECT * FROM Prospects WHERE zip=?")){
$query->bind_param("s", $zip);
$query->execute();
}
echo "<table width='540' cellpadding='0' border='0' cellspacing='0'>
<tr>
<th>Under 4</th>
<th>5 - 9</th>
<th>10 - 14</th>
<th>15 - 17</th>
<th>18 - 20</th>
</tr>";
while($row = mysqli_fetch_array($result))
{
echo "<tr>";
echo "<td>" . $row['cy_pop_04'] . "</td>";
echo "<td>" . $row['cy_pop_59'] . "</td>";
echo "<td>" . $row['cy_pop_1014'] . "</td>";
echo "<td>" . $row['cy_pop_1517'] . "</td>";
echo "<td>" . $row['cy_pop_1820'] . "</td>";
echo "</tr>";
}
echo "</table>";
} // closing brace for if(isset($_POST['submit']))
mysqli_close($con);
?>
Footnotes:
Do not do or use this:
WHERE zip = 'echo $_POST['addressInput']'
^^^^ ^ ^
It's always better using prepared statements when using mysqli_* functions.
Here is a tutorial on using prepared statements.
I am having problem in getting values from db. Iam new in php
I am using checkboxes to get values from database. Only checked values should be printed.
<form method="POST" action="gradoviexport.php" id="searchform">
<div id="GRADOVI BIH">
<h3>GRADOVI BOSNE I HERCEGOVINE</h3><hr/>
<input type="checkbox" name="gradovi[]" value="sarajevo"> Sarajevo
<input type="checkbox" name="gradovi[]" value="banovici"> Banovići
<input type="checkbox" name="gradovi[]" value="banjaluka"> Banja Luka
<input type="checkbox" name="gradovi[]" value="bihac"> Bihać
<input type="checkbox" name="gradovi[]" value="bileca"> Bileća
</div>
<div id="snimi">
<input type="submit" name="submit" value="EXPORT">
</div>
</form>
If Sarajevo is checked I want to print values from database. It does not have to be only one value checked If all values are checked it should print all values.
$con=mysqli_connect("$host","$username","$password", "$database");
// Check connection
if (mysqli_connect_errno())
{
echo "Failed to connect to MySQL: " . mysqli_connect_error();
}
//connecting to db
$variable=$_POST['grad'];
foreach ($variable as $variablename)
{
$sql_select="SELECT * FROM `clanovi` WHERE `GRAD` = $variablename " ;
$queryRes = mysql_query($sql_select);
print"$sql_select";
}
echo "<table border='5'>
<tr>
<th>IME</th>
<th>PREZIME</th>
<th>FIRMA</th>
<th>ADRESA</th>
<th>TELEFON</th>
<th>FAX</th>
<th>MOBITEL</th>
<th>EMAIL </th>
<th>WEB_STRANICA </th>
<th>GRAD </th>
<th>KATEGORIJA </th>
</tr>";
while($row = mysqli_fetch_array($queryRes))
{
echo "<tr>";
echo "<td>" . $row['IME'] . "</td>";
echo "<td>" . $row['PREZIME'] . "</td>";
echo "<td>" . $row['FIRMA'] . "</td>";
echo "<td>" . $row['ADRESA'] . "</td>";
echo "<td>" . $row['TELEFON'] . "</td>";
echo "<td>" . $row['FAX'] . "</td>";
echo "<td>" . $row['MOBITEL'] . "</td>";
echo "<td>" . $row['EMAIL'] . "</td>";
echo "<td>" . $row['WEB_STRANICA'] . "</td>";
echo "<td>" . $row['GRAD'] . "</td>";
echo "<td>" . $row['KATEGORIJA'] . "</td>";
echo "</tr>";
}
echo "</table>";
mysqli_close($con);
Assume you posted gradovi[] array values to submitted page.
Submit page:
$grad = array();
$grad = $_POST['gradovi']; //get array value
$grad = implode(',',$grad); //convert it into comma separated string
//Insert it into data base
Getting from database:
//fetch the gradovi field from the db like below
echo $row['gradovi']; // print all values
or
$grad = explode(',',$row['gradovi']);
foreach($grad as $check) {
echo $check; //print one by one
}
There is few errors in your code.
There is no escaping of the string from POST data. Use mysqli_real_escape_string
There is an error in your while loop. You redefining mysql query result.
Fixed code:
//connecting to db
$variable=$_POST['grad'];
foreach($variable as $key => $val) {
$variable[$key] = mysql_escape_string($val);
}
$sql_select="SELECT * FROM `clanovi` WHERE `GRAD` IN ('" . implode("','", $variable) . "')" ;
$queryRes = mysql_query($sql_select);
print"$sql_select";
I have been batting this around for a while and can not get the variables working for a search. Can connect fine and return results defined as a proper mssql_query and also am fine order by on variables etc but trying to get a text search implemented is not returning results not sure why.
<?php
$link = mssql_connect('SERV13\\RALSQL12', 'RA4joomla', 'Fenestron1');
if (!$link || !mssql_select_db('RALNHV', $link)) {
die('Unable to connect or select database!');
}else{
echo"";
}
if(isset($_REQUEST['submit'])){
$firstname=$_POST['FirstName'];
$surname=$_POST['Surname'];
$query = 'SELECT * FROM lEmployee WHERE FirstName LIKE '%".$firstname."%' OR Surname LIKE '%".$surname."%'';
$q=mssql_query($sql);
}
else{
$query = 'SELECT * FROM lEmployee';
$q = mssql_query($query);
}
?>
<form method="post">
<table width="200" border="1">
<tr>
<td>Name</td>
<td><input type="text" name="firstname" value="<?php echo $firstname;?>" /></td>
<td>Email</td>
<td><input type="text" name="surname" value="<?php echo $surname;?>" /></td>
<td><input type="submit" name="submit" value=" Find " /></td>
</tr>
</table>
</form>
<?php
// Check if there were any records
echo "<table class='table'>";
echo "<tr>";
echo "<th><a href='?orderBy=FirstName'>FirstName</a></th><th><a href='?orderBy=Surname'>Surname</a></th><th><a href='?orderBy=EmployeeNo'>Trigram</a></th><th>Office Phone</th><th>Mobile</th><th><a href='?orderBy=EmployeeJobTitle'>Job Title</a></th><th><a href='?orderBy=Name'>Base</a></th>";
echo "</tr>";
while ($row = mssql_fetch_array($query)) {
echo "<tr>";
echo "<td>" . '' . iconv("CP1252", "UTF-8", $row['FirstName']) . '' . "</td>";
echo "<td>" . '' . iconv("CP1252", "UTF-8", $row['Surname']) . '' . "</td>";
echo "<td>" . '' . $row['EmployeeNo'] . '' . "</td>";
echo "<td>" . '' . $row['Phone'] . '' . "</td>";
echo "<td>" . '' . $row['Mobile'] . '' . "</td>";
echo "<td>" . '' . $row['EmployeeJobTitle'] . '' . "</td>";
echo "<td>" . '' . $row['Name'] . '' . "</td>";
echo "</tr>";
}
echo "</table>";
?>
I am 100% agree with nickL you have some formating issue in your query try to replace your search query by this:
$firstname=$_POST['firstname'];
$surname=$_POST['surname'];
$query = "SELECT * FROM lEmployee WHERE FirstName LIKE '%".$firstname."%' OR Surname LIKE '%".$surname."%'";
$q=mssql_query($sql);
php is a case sensitive language your post variables name are wrong replace the code and try again, if not succeeded try echo $query and run it in query browser in sql server.
hope this will fix the issue.