I'm having trouble understanding the documentation for the UPDATE command of MYSQL. I am viewing records in a PHP page from the database and I want to edit them.
To INSERT I have this code which is an array. I want to know if I can do the same with the UPDATE statement, to save me lots of this=$this 's.
Insert
mysql_query("INSERT INTO $tbl_name(title, pbDate, summary, blog) VALUES('$title', 'pbDate', '$summary', '$blog')")or die(mysql_error());
Update
mysql_query("UPDATE $tbl_name SET title='$title', pbDate='$pbDate' summary='$summary' blog='$blog' WHERE id='$id'")
I thinking something like this, but I'm not sure and can't find anything in the manual.
mysql_query("UPDATE $tbl_name SET (title, pbDate, summary, blog) VALUES('$title', 'pbDate', '$summary', '$blog') WHERE id='$id'")
You could use an array ...
Working phpFiddle:
http://phpfiddle.org/main/code/pi9-ckh
<?php
$array = array(
"column" => "some_value",
"title" => "some_title",
);
$setString = '';
foreach($array as $key => $value){
$tempArray[] = $key . ' = ' . "\"" . $value . "\"";
}
$setString = implode(",", $tempArray);
echo $setString;
?>
Related
Suppose I have following data :
a:1:{s:4:"date";a:3:{s:2:"mm";s:2:"12";s:2:"dd";s:2:"06";s:2:"yy";s:4:"1991";}}
I want the value of mm and dd from that serialized string.
I have following query to retrieve above data :
$birthday = "select meta_key, meta_value from $wpdb->usermeta where meta_key='pie_date_5'";
$user_birthday = $wpdb->get_results($birthday);
foreach($user_birthday as $ubday){
$ubd = $ubday->meta_value; echo $ubd;
echo json_decode($ubd[0]->mm);
}
I am not able to do that simple thing. Please help me out.
You have to use unserialize instead of json_decode because it's a serialized string.
So you could try something like this:
$user_birthday = 'a:1:{s:4:"date";a:3:{s:2:"mm";s:2:"12";s:2:"dd";s:2:"06";s:2:"yy";s:4:"1991";}}'; //the serialized string you get from your query
$upday = unserialize( $user_birthday );
echo $upday['date']['mm'] . PHP_EOL;
echo $upday['date']['dd'];
EDIT: If you want to read all properties from the $upday array you could use a foreach-loop
foreach ( $upday['date'] as $key => $value ) :
echo $key . ' = ' . $value . PHP_EOL;
endforeach;
I am trying to export my MySQL tables from my database to a JSON file, so I can list them in an array.
I can create files with this code no problem:
$sql=mysql_query("select * from food_breakfast");
while($row=mysql_fetch_assoc($sql))
{
$ID=$row['ID'];
$Consumption=$row['Consumption'];
$Subline=$row['Subline'];
$Price=$row['Price'];
$visible=$row['visible'];
$posts[] = array('ID'=> $ID, 'Consumption'=> $Consumption, 'Subline'=> $Subline, 'Price'=> $Price, 'visible'=> $visible);
}
$response['posts'] = $posts;
$fp = fopen('results.json', 'w');
fwrite($fp, json_encode($response));
fclose($fp);
Now this reads a table and draws it's info from the fields inside it.
I would like to know if it is possible to make a JSON file with the names of the tables, so one level higher in the hierarchy.
I have part of the code:
$showtablequery = "
SHOW TABLES
FROM
[database]
LIKE
'%food_%'
";
$sql=mysql_query($showtablequery);
while($row=mysql_fetch_array($sql))
{
$tablename = $row[0];
$posts[] = array('tablename'=> $tablename);
}
$response['posts'] = $posts;
But now i am stuck in the last line where is says: $ID=$row['ID']; This relates to the info inside the Table and I do not know what to put here.
Also as you can see, I need to filter the Tables to only list the tables starting with food_ and drinks_
Any help is greatly appreciated:-)
There is no 'table id' in MySQL and therefore the result set from SHOW TABLES has no index id. The only index in the resultset is named 'Tables_in_DATABASENAME'.
Also you should use the mysqli library as the good old mysql library is depreacted. Having prepared an example:
<?php
$mysqli = new mysqli(
'yourserver',
'yourusername',
'yourpassword',
'yourdatabasename'
);
if ($mysqli->connect_errno) {
echo "Failed to connect to MySQL: (" . $mysqli->connect_errno . ") "
. $mysqli->connect_error;
}
$result = $mysqli->query('SHOW TABLES FROM `yourdatabasename` LIKE \'%food_%\'');
if(!$result) {
die('Database error: ' . $mysqli->error);
}
$posts = array();
// use fetch_array instead of fetch_assoc as the column
while($row = $result->fetch_array()) {
$tablename = $row[0];
$posts []= array (
'tablename' => $tablename
);
}
var_dump($posts);
Please help to create a query!
This is a working example of my query Retrieve data as JSON using PHP:
<?php
$con = mysql_connect("localhost","user","password");
if (!$con) {
die('Could not connect: ' . mysql_error());
}
mysql_select_db("admin_accounting", $con);
$result = mysql_query("SELECT unix_timestamp(date), sum(ksi2k) FROM accounting where lhc_vo like 'ops' group by year(date), month(date)");
$rows = array();
$rows['type'] = 'area';
$rows['name'] = 'Ops';
while($r = mysql_fetch_array($result)) {
$rows['data'][] = $r[0]*1000;
$rows['data'][] = $r[1];
array_push($rows);
}
print json_encode($rows, JSON_NUMERIC_CHECK);
mysql_close($con);
?>
The JSON results look like this:
{"type":"area","name":"Ops","data":[1167664515000,0,1170342915000,0,1172762115000,0,1175436915000,0,1178028915000,0]}
But I need the JSON results should look like this:
{"type":"area","name":"Ops","data":[[1167664515000,0],[1170342915000,0],[1172762115000,0],[1175436915000,0],[1178028915000,0]]}
I would be very grateful for the help
while($r = mysql_fetch_array($result)) {
$rows['data'][] = array($r[0]*1000, $r[1]);
}
In addition to my other answer, you ought to consider switching away from the ancient (and now deprecated, as of PHP v5.5) ext/mysql. Here is an example using PDO, in which you can see how simple your problem becomes:
<?php
$con = new PDO(
'mysql:hostname=localhost;dbname=admin_accounting',
'user',
'password'
);
$result = $con->query('
SELECT 1000*UNIX_TIMESTAMP(date), SUM(ksi2k)
FROM accounting
WHERE lhc_vo LIKE "ops"
GROUP BY YEAR(date), MONTH(date)
');
print json_encode([
'type' => 'area',
'name' => 'Ops',
'data' => $result->fetchAll(PDO::FETCH_NUM)
]);
?>
Note that:
I have moved the multiplication into the database layer in order that I can simply call fetchAll() to obtain the resulting array;
MySQL will select an indeterminate value from amongst those in each group for the first column in the resultset; should this be undesirable, you will need to apply a suitable aggregate function to the reference to the date column; and
I have used the short array syntax, which is only available from PHP v5.4—if you're using an earlier version, you will need to replace the [ … ] of the argument to json_encode() with array( … ).
I want to thank everyone here for the help I have recieved so far. My next question is a bit more complicated.
So I have a database set up on my server, and I have a form on my website where I am submitting data to my MYSQL database.
After I submit the data, I am having trouble searching for it, displaying possible results, and then making those results HYPERLINKED so that the user can find out more about they are looking for.
My "common.php" script is set up like this:
<?php
$username = "XXX";
$password = "XXX";
$hostname = "XXX";
$database = "XXX";
mysql_connect($hostname, $username, $password, $database) or die
("Unable to connect to MySQL");
echo "Connected to MySQL<br>";
?>��
My "insertdata.php" script is set up like this:
<?php
require("common.php");
// connect with form
$name=$_POST['firstname'];
$lastname=$_POST['lastname'];
$city=$_POST['city'];
$state=$_POST['state'];
$zip=$_POST['zip'];
$phone=$_POST['phone'];
$email=$_POST['email'];
$various=$_POST['various'];
$other=$_POST['other'];
// insert data into mysql
$query="INSERT INTO datatable
(
firstname,
lastname,
city,
state,
zip,
phone,
email,
various,
other,
)
VALUES
(
'$firstname',
'$lastname',
'$city',
'$state',
'$zip',
'$phone',
'$email',
'$various',
'$other',
)";
$result=mysql_query($query);
// if successfull displays message "Data was successfully inserted into the database".
if($result){
echo "Successful";
echo "<BR>";
echo "<a href='insert.php'>Back to main page</a>";
}
else {
echo "ERROR... data was not successfully insert into the database";
}
mysql_close();
?>��
From there, I want to make the inserted data searchable.
My problem is, when the search is completed, I want to only display the First Name and Last Name in two separate columns.
From there, I want a link displayed in a third separate column with a link in each row that says "View Record Details."
Finally, when "View Record Details" in clicked, it brings me to the correct record, formatted again in an HTML table.
The closest I have come to a solution is:
<?php
require("common.php");
$query="SELECT * FROM datatable";
$result=mysql_query($query);
$num=mysql_numrows($result);
$i=0;
while ($i < $num) {
$firstname=mysql_result($result,$i,"firstame");
$lastname=mysql_result($result,$i,"lastname");
$i++;}
?>
As an additional question, when I use PDO, does that change my HTML?
Switch to PDO. Your code will look something like this:
$conn = new PDO('mysql:host=db_host;dbname=test', $user, $pass);
$sql = 'SELECT * FROM datatable';
foreach ($conn->query($sql) as $row) {
print $row['firstname'] . "\t";
print $row['lastname'] . "\n";
}
EDIT:
To link back for details add this line after the 2nd print:
print "<a href='somephp.php?idx=" . $row[ 'idx' ] . "'>link here</a>";
You'll need another php file called 'somephp.php':
$conn = new PDO('mysql:host=db_host;dbname=test', $user, $pass);
$idx = $_REQUEST[ 'idx' ];
$sql = 'SELECT * FROM datatable where idx = ?';
$stmt = $conn->prepare( $sql );
$stmt->bindParam( 1, $idx );
$stmt->execute();
$row = $stmt->fetch();
// now print all the values...
print $row['firstname'] . "\t";
print $row['lastname'] . "\t";
print $row['address'] . "\t";
and so on...
NOTE: This depends on each record having a unique key 'idx'. I don't see this in your values above so you'll have to find a way to incorporate it if you want to use this code.
ALSO: You ask - does this change the HTML and does this handle table formatting - No to both. You do all the HTML formatting via the print statements. All PHP does it output lines to the browser.
I am a newbie in Drupal, I have a table in a drupal database. I wanted to query all the content of it and display in a tabular format. Whenever I execute the query the link directs me to the blank page and nothing appears. I have attached the PhP code for the reference. Any help will be highly appreciated.
<?php
$header = array('Name', 'Age', 'Sex','University');
$rows = array();
$sql = 'SELECT Name, Age, Sex,University FROM {data_pulling} ORDER BY Name';
$res = db_query($sql);
while ($row = db_fetch_array($res)) {
$rows[] = $row;
}
print theme('table', $header, $rows);
?>
At a guess you're using Drupal 7 but are trying to use API functions from Drupal 6.
Try this
foreach ($res as $row) {
$rows[] = (array) $row;
}
print theme('table', array('rows' => $rows, 'header' => $header));
Have a look at the docs for db_query and theme_table for more information