Mysql select and insert using prepared statement - mysql

Question

My mysql query is working fine
INSERT INTO donor_location (pc_id)
SELECT id
FROM pc
WHERE postcode= ?
i.e gets the postcode id from a postcode table then inserts that id into donor_location table.
I am using mysqli and prepared statements
without the select part it would be quite easy - something like
$stmt = $mysqli->prepare("INSERT INTO donor_charity(
id) values (?)") ;
however I am completely lost about how to incorporate the select

Answer 1

What you do is almost the same, just changing the query bit.
To select all records from charity_donor where the id is 25, you would do the follwing query:
SELECT *
FROM donor_charity
WHERE id = 25
Now to query this, first you have to prepare it:
$stmt = $mysqli->prepare("
SELECT *
FROM donor_charity
WHERE id = ?
");
Now to loop over the results, you must bind the param, and execute the query.
$stmt->bind_param('d', 25 ); // First param means the type of the value you're
passing. In this example, d for digit.
$stmt->execute();
Then you setup an array to hold the data returned from the query,
$row = array();
stmt_bind_assoc($stmt, $row);
And now to loop over the returned data.
while ( $stmt->fetch () ) {
print_r($row); // Should now contain the column.
}
For documentation, see:
Prepare: http://www.php.net/manual/en/mysqli.prepare.php
Bind param: http://www.php.net/manual/en/mysqli-stmt.bind-param.php
Execute: http://www.php.net/manual/en/mysqli-stmt.execute.php
Fetch: http://www.php.net/manual/en/mysqli-stmt.fetch.php

Answer 2

You need to use Bind_param after Prepare statement.
$sql = "INSERT INTO donor_charity(
id) values (?)
";
/* create a prepared statement */
if (!$stmt = $db->prepare($sql)) {
echo 'Database prepare error';
exit;
}
/* bind parameters for markers */
$stmt->bind_param('ssssss', $id);
$id = '123456';
/* execute query */
$stmt->execute();

Answer 3

Hope this post helps, it's so simple.
http://www.java2s.com/Code/Java/Database-SQL-JDBC/InsertRecordsUsingPreparedStatement.htm