Please advise me if I haven't written a program about the Mysqli prepare error for a long time. due to the many advantages of the mysqli class database upgraded by php5.0, I want to use the mysqli method to operate the database, after checking materials and reading manuals on the Internet, browsing and deleting can be done normally, but the insertion function always prompts an error: Fatal error: Call to a member function prepare () on null in D: \ xampps \ htdocs \ txl \ addbook. php on line 20.
I am dying. can't I solve this problem? do I have to give up Mysqli? Please give me some advice on where my code is wrong? Or a server problem? (The server I use is the Xampps1.9.7 Integrated Environment)
If ($ _ POST ['contacts'] = "true ")
{
$ Stmt = $ mysqli-> prepare ("insert into contactsuser (bookUname, bookDep, bookDuty, bookphone, bookTel, bookEmail, bookGid) VALUES (?, ?, ?, ?, ?, ?, ?) ");
$ Stmt-> bind_param ('sssd', $ bookUname, $ bookDep, $ bookDuty, $ bookphone, $ bookTel, $ bookEmail, $ bookGid );
$ BookUname = $ _ POST ['bookuname'];
$ BookDep = $ _ POST ['bookdep'];
$ BookDuty = $ _ POST ['bookduty'];
$ Bookphone = $ _ POST ['bookphone'];
$ BookTel = $ _ POST ['booktel'];
$ BookEmail = $ _ POST ['bookemail'];
$ BookGid = $ _ POST ['bookgid'];
/* Execute prepared statement */
$ Stmt-> execute ();
Printf ("% d Row inserted. \ n", $ stmt-> affected_rows );
/* Close statement and connection */
$ Stmt-> close ();
Exit;
}
Reply to discussion (solution)
$ Mysqli is not a mysqli object
Try the process-oriented method:
$stmt = mysqli_prepare($link, "INSERT INTO contactsuser(bookUname, bookDep, bookDuty,bookphone,bookTel,bookEmail,bookGid) VALUES (?, ?, ?, ?, ?, ?, ?)")
I did not forget to include the database link file in the program just now, but there are still errors after the inclusion, including the change to the process-oriented style:
Warning: mysqli_stmt_bind_param (): Number of elements in type definition string doesn't match number of bind variables in D: \ xampps \ htdocs \ txl \ addbook. php on line 24
The procedure is as follows:
$ Link = mysqli_connect ('localhost', 'root', 'root', 'ssssbook ') or die ('unale to connection ');
$ Stmt = mysqli_prepare ($ link, "insert into contactsuser (bookUname, bookDep, bookDuty, bookphone, bookTel, bookEmail, bookGid) VALUES (?, ?, ?, ?, ?, ?, ?) ");
Mysqli_stmt_bind_param ($ stmt, 'sssd', $ bookUname, $ bookDep, $ bookDuty, $ bookphone, $ bookTel, $ bookEmail, $ bookGid );
$ BookUname = $ _ POST ['bookuname'];
$ BookDep = $ _ POST ['bookdep'];
$ BookDuty = $ _ POST ['bookduty'];
$ Bookphone = $ _ POST ['bookphone'];
$ BookTel = $ _ POST ['booktel'];
$ BookEmail = $ _ POST ['bookemail'];
$ BookGid = 1;
Mysqli_stmt_execute ($ stmt );
Mysqli_stmt_close ($ stmt );
I also checked the field type of the database. except that the primary key bookid is the automatically added int type, the bookGid is the int type, and the other is the varchar type.
VALUES (?, ?, ?, ?, ?, ?, ?)
$ Stmt, 'sssd', $ bookUname, $ bookDep, $ bookDuty, $ bookphone, $ bookTel, $ bookEmail, $ bookGid
The seven question marks correspond to eight parameters ..... Add a question mark and try again.
mysqli_stmt_bind_param($stmt,'ssssssd', $bookUname, $bookDep, $bookDuty, $bookphone, $bookTel, $bookEmail, $bookGid);
Http://php.net/manual/zh/mysqli-stmt.bind-param.php
The number of variables and length of string types must match the parameters in the statement.
The upstairs is correct. it should be ssssssd.
Great! Thank you for your help! In particular, the question "ssssssd" mentioned by saint_leer and misakaqunianx... has been mentioned, and I have never understood what this character string means!