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I´m trying to insert a current SELECT-Statement into an Database table with an PHP-variable. It works by this way without the variable:
mysql_query("INSERT INTO table (column) SELECT column FROM table1");
Did anyone have a idea how I can fix it?
PHP: 5.3.10 and mysql-databse
try this
$sample="SELECT column FROM table1"; // your select query
$query="INSERT INTO `table` (column) $sample";
mysqli_query('database connection code',$query);
Let me know if any problem arises.
You can use mysql_insert_id() to get the last inserted ID value and try insert query separately.
<?PHP
$lastId = mysql_insert_id();
$query = mysql_query("Your Select Query with $lastId");
?>
Related
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The image above is my sql database. what I want to do is load the names and echo in json but the thing is though, I want to show one of each name. Like see how there are 4 assassinshadow entries? I want php to echo only one of em. not making much sense am I? haha
You can use DISTINCT in your mysql query:
$mysqli = mysqli_connect("example.com", "user", "password", "database");
$query = "SELECT DISTINCT name from yourtable";
$res = mysqli_query($mysqli, $query);
$row = mysqli_fetch_assoc($res);
Two ways to do that:
SELECT distinct name FROM my_table
or
SELECT name FROM my_table group by name
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In my MySQLi database, I have 2 columns in the table. One is 'author' and the other is 'books'. I have a variable called '$author' and an array called '$books' which contains unknown numbers of values.
I want to populate column 'books' with values inside array '$books' and in the other column variable '$author' which will remain constant in all rows.
Please help. It will be even more appreciated if you provide it in procedural way instead of OOP.
Assuming $author is a string and $books is an array of integers,
$sql = "INSERT INTO your_table (books,author) VALUES(";
foreach($books as $book_data){
$sql .= "($book_data,'$author'),";
}
$sql = rtrim($sql, ",") . ")";
//Execute the query
This will insert every $books variable as a new row with the constant $author.
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so I have records in my database that I would like to print to a HTML table IF they belong to the user.
$sql="SELECT * FROM ".$tbl_name." WHERE Username='".$username."'";
The table would read :
Date, Room, Period , Cancel
With cancel being a red cross designed to delete that specific entry from the database :
Booking ID, Date, Period,Type,RoomID, Username are the fields in the database.
How would I go about doing this? I was thinking echoing a table using a for loop but I wouldn't know where to begin?
Try:
$query_rsSearch = "SELECT * FROM ".$tbl_name." WHERE Username='".$username."'";
$rsSearch = mysql_query($query_rsSearch) or die(mysql_error());
$row_rsSearch = mysql_fetch_assoc($rsSearch);
$totalRows_rsSearch = mysql_num_rows($rsSearch);
do {
echo $row_rsSearch['Date']." | ".$row_rsSearch['Room']." | ".$row_rsSearch['Period'];
} while($row_rsSearch = mysql_fetch_assoc($rsSearch));
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This statement works in pgAdmin but not when run in a php script the php script can select all but can not update why is this?
UPDATE users SET password = '123123' WHERE email = 'random#random.com'
PHP code that doesn't work:
$sql = $dbh->prepare("UPDATE users SET password = '11111111111' WHERE email = 'test#outlook.com')");
$sql->execute(array());
PHP code that does work:
$sql = $dbh->prepare("SELECT * FROM users");
$sql->execute(array());
$fr = $sql->fetchAll(); var_dump($fr);
In your update query you've got ) at the end which will cause syntax error. Check it using eg. $dbh->errorInfo().
Also, don't use prepare() for queries that don't use parameters. Instead use query() for SELECT and exec() for others.
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This is the site that I am trying to create the script for http://facechat.louisdickinson.com/
The idea is people can save there email to my database, and another button called "Start Call" will randomly select a email and call it using:
facetime://email#email.com
Effectively this will create a "omegle" style web-based facetime chat site.
I am new to MySQL and PHP and don't know where to start, any help will be appreciated!
I don't know, what kind of help do you need.. First of all, you should have a php script, witch can take the posted name/e-mail pair.
In this script, you should sanitize the posted values, than you can add it to your database with the following:
$query = "INSERT INTO <tableName> (`name`, `e-mail`) VALUES ( '".$postedName."', '".$postedMail."' )";
On button press, you should have another php script, for selecting the random e-mail:
$query = "SELECT COUNT(*) FROM <tableName>";
$max should be the query result.
$random = rand( 0 , $max - 1 );
$query = "SELECT `e-mail` FROM <tableName>" LIMIT $random, 1";
With this query you got a random e-mail.
Do you need more exact code? Please be more exact on what you need!
Kind regards,
hotzu
In php:
mysql_connect("host", "user", "password") or die(mysql_error());
mysql_selectdb("database"); // '' for an auto increment column
mysql_query("INSERT INTO TABLE VALUES ('$email', ''));
And for more info http://www.w3schools.com/php/php_mysql_intro.asp