I want to echo onto my screen all the data is my 'userdata' table. i have looked around and found this code but when i run it i get a HTTP ERROR 500.
this is my code that im trying to use:
<?php
$database = new SQLite3('home.db');
$result = $database->query("SELECT * FROM userdata");
echo $result;
?>
the $database->query() method will return an SQLite3Result object which you can't just "echo". Instead, you should loop through all the results like so:
<?php
$database = new SQLite3('home.db');
$result = $database->query("SELECT * FROM userdata");
while ($row = $result->fetchArray()) {
print_r($row);
}
?>
The $row variable inside the while loop will be an array. Use the appropriate index to get the value of a single column if necessary.
Related
I am trying to store the result from a MySQL statement for later use in PHP. I have this code which gets me the result:
// Get the categories from the db.
$categories = array();
$catSql = "SELECT id, name FROM categories";
if ($catStmt = mysqli_prepare($db, $catSql))
{
$catStmt->execute();
$result = $catStmt->get_result();
// Fetch the result variables.
while ($row = $result->fetch_assoc())
{
// Store the results for later use.
}
}
So I know i will have the results in $row["id"] and $row["name"] and I want to save all of the rows so whenever i need them i can loop through them and for example echo them. I have searched for structures and arrays for keeping them in PHP but I cannot seem to find any information about that or maybe I am not searching in the right direction. Can anyone point me where i should read about this to find out how to do this efficiently and if possible post a small example?
Use sessions:
if (session_status() == PHP_SESSION_NONE) {
session_start();
}
// Get the categories from the db.
$categories = array();
$catSql = "SELECT id, name FROM categories";
if ($catStmt = mysqli_prepare($db, $catSql))
{
$catStmt->execute();
$result = $catStmt->get_result();
// Fetch the result variables.
while ($row = $result->fetch_assoc())
{
// Store the results for later use.
$_SESSION['category_' . $row['id']] = $row['name'];
}
}
Then access it later from a different page
$_SESSION['session_variable_name']
You can also create an array of the information and store the entire array in a single session variable.
Just make sure you add the session_start function at the beginning of each page. The if statement prevents you from trying to start it multiple times.
$categories = array();
$catSql = "SELECT id, name FROM categories";
if ($catStmt = mysqli_prepare($db, $catSql))
{
$catStmt->execute();
$result = $catStmt->get_result();
while ($row = $result->fetch_assoc())
{
$categories[$row['id']]=$row['name'];
}
}
And If you want the name anywhere use below :
$categories[$id]
I create as the following function. how to get all data using this array. when run this function will appear only the first record. but, i want it to appear all the records. what is the error in this code.
public function get_All_Se($stId){
$query = "SELECT * FROM session WHERE stId = '$stId'";
$result = $this->db->query($query) or die($this->db->error);
$data = $result->fetch_array(MYSQLI_ASSOC);
return $data;
}
public function get_All_Se($stId){
$rows=array();
$query = "SELECT * FROM session WHERE stId = '$stId'";
$result = $this->db->query($query) or die($this->db->error);
while($data= $result->fetch_assoc()){
$rows[]=$data;
}
return $rows;
}
Run loop over all results and add to some return array.
$rows = array();
while(($row = $result->fetch_array($result))) {
$rows[] = $row;
}
As the documentation of mysqli::fetch_array() explains, it returns only one row (and not an array containing all the rows as you might think).
The function you are looking for is mysqli::fetch_all(). It returns all the rows in an array.
public function get_All_Se($stId)
{
$query = "SELECT * FROM session WHERE stId = '$stId'";
$result = $this->db->query($query) or die($this->db->error);
return $result->fetch_all(MYSQLI_ASSOC);
}
The code above still has two big issues:
It is open to SQL injection. Use prepared statements to avoid it.
or die() is not the proper way to handle the errors. It looks nice in a tutorial but in production code it is a sign you don't care about how your code works and, by extension, what value it provides to their users. Throw an exception, catch it and handle it (log the error, put some message on screen etc) in the main program.
Try this way...
<?php
// run query
$query = mysql_query("SELECT * FROM <tableName>");
// set array
$array = array();
// look through query
while($row = mysql_fetch_assoc($query)){
// add each row returned into an array
$array[] = $row;
// OR just echo the data:
echo $row['<fieldName>']; // etc
}
// debug:
print_r($array); // show all array data
echo $array[0]['<fieldName>'];
?>
I have set up a MySql database containing several tables (needed for uploading tons of data that is entered in excel). I would like to use each table name as a variables that automatically updates each time a new table is added to the database. However, I don't know how to prevent PHP from overwriting my $tableName variable in loops. This is what I got so far:
<?php
if (!mysql_connect($host, $username, $password)) {
echo 'Could not connect to mysql';
exit;
}
$sql = "SHOW TABLES FROM $database";
$result = mysql_query($sql);
if (!$result) {
echo "DB Error, could not list tables\n";
echo 'MySQL Error: ' . mysql_error();
exit;
}
while ($row = mysql_fetch_row($result)) {
$tableName = $row[0];
include('../design.php');
}
?>
In other words: On the frontpage of the website I would like to include data from every single database table. I have organized a framework for extracting that data and presenting it neatly in design.php. However, this file makes use of the $tableName variable for extracting the data and I think that variable is being overwritten at each loop causing the error (only the number one table will be displayed).
Instead, if I just separate the code in their own php objects, then everything works just fine:
<?php
$tableName = tablename1;
include('../design.php');
?>
<?php
$tableName = tablename2;
include('../design.php');
?>
etc..
Can anyone explain to me how this works? Or if there is some smarter way of doing things?
Thanks in advance!
include the design.php file at the beginning of the script. Inside this file, define a function.
function doSomethingWithTableName ($table)
{
// do something with it
}
Then, in your frontend file, you would do this:
$sql = "...";
$result = mysql_query ($sql);
while ($row = mysql_fetch_row($result))
{
$processed_data = doSomethingWithTableName ($row[0]);
// do something with $processed_data
}
Hope this helps.
I've no idea what's inside design.php but instead of including it inside the loop I'd create a function, something like:
function designFunction($tableName){
//put the design.php here
return $something;
}
Now use the designFunction inside your loop
while ($row = mysql_fetch_row($result)) {
$tableName = $row[0];
$someResult = designFunction($tableName);
//etc...
}
Just trying to pass a variable on URL so that when echoed I can click on it and open it's own content based on the database record. Right now this one shows all the records from database but what I was trying to do was pass a URL so each blog IDs will have it's own URL and when clicked on it will open the individual entries rather than all the entries.
Edited Now I'm able to show rows of entries with IDs where 'IDs' has URL variable at the end. Do I need to create another query to echo the individual entry on my mini blog?
<?
$db = // connection to db and authentication to connecting to db;
#$postID = $_GET['postID']; // I'm thinking to use a $_GET global variable to work with URL variable
$command = "select * from $table_name"; // I'm thinking to add the Id here or something or create another query to echo the linked URL 'viewblog.php?postID=$data->blogID'
$result = $db->query($command);
while ($data = $result->fetch_object()) {
echo "<TR><TD><a href='viewblog.php?postID=$data->blogID'>".$data->blogID."</a></TD>";
echo "<TD>".$data->author."</TD>";
echo "<TD>".$data->date."</TD>";
echo "<TD>".$data->entry."</TD></TR>\n";
}
$result->free();
$db->close;
Why this script is giving all entries?
Because the final query that is being sent to the database is something like
select * from TABLE_NAME
which will return all entries since your are using the asterix * after SELECT
What you are asking for can be obtained if the executed final query contains the "blogID" before retrieving the results and start fetching them.
http://www.w3schools.com/sql/sql_where.asp
You should also use the fetched or post ID in the echoed result (so that when clicked, each blog has its own id in the link).
It could be something like this
$postID = $_GET['postID'];
//Add filtering by id to select statement
$command = "select * from '$table_name' obj WHERE obj.blogID = '$postID'";
$result = $db->query($command);
while($data = $result->fetch_assoc()){
$data['blogID'] = $postID;
//Add ID to echoed link
echo "<TR><TD> Some Blog (ID: ".$data['blogID'].") </TD>";
echo "<TD>".$data['author']."</TD>";
echo "<TD>".$data['date']."</TD>";
echo "<TD>".$data['entry']."</TD></TR>\n";
}
WATCH OUT for security issues regarding this code. You should use a safer way to do this. I'm only explaining the results.
As for Auto Increment, it can be initiated when you first created the table. This is for when you INSERT a new row into the database. When you use Auto Increment, you don't have to give an ID manually.
http://www.w3schools.com/sql/sql_autoincrement.asp
Notice : The HTML BR ELEMENT should not be used inside TABLE structures.
Hope it helps.
You could create some function like this for returning single post based on url
function single_blog($Post_id){
$sql = "SELECT * FROM your_table WHERE post_id = ? LIMIT 1";
$stmt = $this->db->prepare($sql);
$stmt->execute(array($Post_id);
return $stmt->fetch();
}
You are selecting all entries from your table. Use the following:
$db = // connection to db and authentication to connecting to db;
$postID = $_GET['postID']; // ??
$db->real_escape_string(trim($postID));
$command = "select * from $table_name WHERE `postID`=$postID";
$result = $db->query($command);
// Ensure results before outputting
if ($result->num_rows) while($data = $result->fetch_assoc()){
$data['blogID'] = $postID;
echo "<TR><TD><a href='viewblog.php?postID='>".$data['blogID']."</a> </TD>"; //??
echo "<TD>".$data['author']."</TD>";
echo "<TD>".$data['date']."<BR></TD>";
echo "<TD>".$data['entry']."</TD></TR>\n";
} else echo "No entry found!";
$result->free();
$db->close;
<?php
//$db connect to database
// Entry form sanitation of $_POST
// Insert PHP file to MySQL
// View all blog posts
$postID = $_GET['postID']; // I guess I should sanitize this as well
if (!empty($postID)) {
$command = "select * from $table_name where blogID = $postID";
$result = $db->query($command);
while ($data = $result->fetch_object()) {
$postID = $data->blogID;
echo "<TR><TD>".$postID."</TD>";
echo "<TD>".$data->author."</TD>";
echo "<TD>".$data->date."</TD>";
echo "<TD>".$data->entry."</TD></TR>\n";
}
$result->free();
}
else {
$command = "select * from $table_name";
$result = $db->query($command);
while ($data = $result->fetch_object()) {
$postID = $data->blogID;
echo "<TR><TD><a href='viewblog.php?postID=$postID'>".$postID."</a></TD>";
echo "<TD>".$data->author."</TD>";
echo "<TD>".$data->date."</TD>";
echo "<TD>".$data->entry."</TD></TR>\n";
}
$result->free();
}
$db->close;
?>
I have a table wherein I need to get all the data in one column/field, but I can't seem to make it work with the code I have below:
$con=mysqli_connect("localhost","root","","database");
$result = mysqli_query($con,"select * from client");
$row = mysqli_fetch_array($result111);
echo $row['name'];
With the code above, it only prints one statement, which happens to be the first value in the table. I have 11 more data in the table and they are not printed with this.
You need to loop through the recordsets .. (A while loop will do) Something like this will help
$con=mysqli_connect("localhost","root","","database");
$result = mysqli_query($con,"select * from client");
while($row = mysqli_fetch_array($result))
{
echo $row['name'];
}
The mysqli_fetch_array() function will return the next element from the array, and it will return false when you have ran out of records. This is how you can use while loops to loop through the data, like so:
while ($record = mysqli_fetch_array($result)) {
// do something with the data...
echo $record['column_name'];
}