prepared-statement

How do I escape a literal question mark ('?') in a JDBC prepared statement

北慕城南 提交于 2020-01-01 08:38:08
问题 I'd like to create a JDBC PreparedStatement like: SELECT URL,LOCATE ( '?', URL ) pos FROM Links WHERE pageId=? ORDER BY pos ASC Where the 1st ? is a literal and the 2nd ? is a parameter. I could use CHAR(63) in place of the '?' but I think the extra function call would slow down the SQL execution. Is there some way to escape that 1st ? ? Edit: The following code tests dkatzel's assertion that the ? character in a string is not considered a marker: public class Test { public static void main

What is proper way to use PreparedStatementCreator of Spring JDBC?

为君一笑 提交于 2020-01-01 04:40:31
问题 As per my understanding the use of PreparedStatement in Java is we can use it multiple times. But I have some confusion using PreparedStatementCreator of Spring JDBC. For example consider following code, public class SpringTest { JdbcTemplate jdbcTemplate; PreparedStatementCreator preparedStatementCreator; ResultSetExtractor<String> resultSetExtractor; public SpringTest() throws SQLException { jdbcTemplate = new JdbcTemplate(OracleUtil.getDataSource()); preparedStatementCreator = new

MySQLi Prepared Statements and Transactions

China☆狼群 提交于 2019-12-31 22:22:15
问题 Is there a way to do transactions with prepared statements? I mean can I use the following example with $mysqli->autocommit(FALSE); and $mysqli->commit( ); and $mysqli->rollback( ); //Preparing the statment $insert_stmt=$mysqli->prepare("INSERT INTO x VALUES(?,?)") or die($mysqli->error); //associate variables with the input parameters $insert_stmt->bind_param("is", $my_number,$my_string); //i=integer //Execute the statement multiple times.... for ($my_number = 1; $my_number <= 10; $my_number

Missing STOPKEY per partition in Oracle plan for paging by local index

冷暖自知 提交于 2019-12-31 10:33:45
问题 There is next partitioned table: CREATE TABLE "ERMB_LOG_TEST_BF"."OUT_SMS"( "TRX_ID" NUMBER(19,0) NOT NULL ENABLE, "CREATE_TS" TIMESTAMP (3) DEFAULT systimestamp NOT NULL ENABLE, /* other fields... */ ) PCTFREE 10 PCTUSED 40 INITRANS 1 MAXTRANS 255 STORAGE(BUFFER_POOL DEFAULT FLASH_CACHE DEFAULT CELL_FLASH_CACHE DEFAULT) TABLESPACE "ERMB_LOG_TEST_BF" PARTITION BY RANGE ("TRX_ID") INTERVAL (281474976710656) (PARTITION "SYS_P1358" VALUES LESS THAN (59109745109237760) SEGMENT CREATION IMMEDIATE

Missing STOPKEY per partition in Oracle plan for paging by local index

柔情痞子 提交于 2019-12-31 10:32:11
问题 There is next partitioned table: CREATE TABLE "ERMB_LOG_TEST_BF"."OUT_SMS"( "TRX_ID" NUMBER(19,0) NOT NULL ENABLE, "CREATE_TS" TIMESTAMP (3) DEFAULT systimestamp NOT NULL ENABLE, /* other fields... */ ) PCTFREE 10 PCTUSED 40 INITRANS 1 MAXTRANS 255 STORAGE(BUFFER_POOL DEFAULT FLASH_CACHE DEFAULT CELL_FLASH_CACHE DEFAULT) TABLESPACE "ERMB_LOG_TEST_BF" PARTITION BY RANGE ("TRX_ID") INTERVAL (281474976710656) (PARTITION "SYS_P1358" VALUES LESS THAN (59109745109237760) SEGMENT CREATION IMMEDIATE

PHP login form returning error

时光总嘲笑我的痴心妄想 提交于 2019-12-31 07:41:49
问题 I have created a login script below, however I have been told to use prepared statements <?php require '../php/connect.php'; if(isset($_POST['login'])){ $username = mysqli_real_escape_string($con, $_POST['username']); $password = mysqli_real_escape_string($con, $_POST['password']); $select_user = "SELECT * FROM users WHERE username = '$username' AND password = '$password'"; $run_user = mysqli_query($con, $select_user); $check_user = mysqli_num_rows($run_user); if($check_user>0){ header(

PHP login form returning error

风流意气都作罢 提交于 2019-12-31 07:41:25
问题 I have created a login script below, however I have been told to use prepared statements <?php require '../php/connect.php'; if(isset($_POST['login'])){ $username = mysqli_real_escape_string($con, $_POST['username']); $password = mysqli_real_escape_string($con, $_POST['password']); $select_user = "SELECT * FROM users WHERE username = '$username' AND password = '$password'"; $run_user = mysqli_query($con, $select_user); $check_user = mysqli_num_rows($run_user); if($check_user>0){ header(

undefined function mysqli_stmt_init() php error

冷暖自知 提交于 2019-12-31 04:45:31
问题 I'm new to using php mysqli prepared statements. No matter what I try I always get this error message. Fatal error: Call to undefined function mysqli_stmt_init() in...(etc) I have close my database link further below in my code, it isn't show here. Here is my code: $link = mysqli_connect($mysql_host, $mysql_user, $mysql_password, $mysql_database); if (mysqli_connect_errno()) { echo 'We\'re having problems connecting right now. Please try again later.'; exit(); } $email_query = mysqli_stmt

Using fetch_assoc on prepared statements (php mysqli)

懵懂的女人 提交于 2019-12-30 18:43:49
问题 I'm currently working on a login script, and I got this code: $selectUser = $db->prepare("SELECT `id`,`password`,`salt` FROM `users` WHERE `username`=?"); $selectUser->bind_param('s', $username); $selectUser->execute(); if ($selectUser->num_rows() < 0) echo "no_user"; else { $user = $selectUser->fetch_assoc(); echo $user['id']; } Here's the error I get: Fatal error: Uncaught Error: Call to undefined method mysqli_stmt::fetch_assoc() I tried all sorts of variations, like: $result = $selectUser

Mysqli Prepared Stmt returns 0 num_rows

↘锁芯ラ 提交于 2019-12-30 11:32:12
问题 Help. I am getting 0 num_rows but if i execute the query in the console i am getting results. I m kinda new to prepared stmts. Here is my code Database connection class: class DbConnection { const HOST = "localhost"; const USR = "root"; const PWD = ""; const DB = "club_db"; } Login class: class UsrLogin extends DbConnection { private $conn; /*db connector*/ /*login vars*/ private $usr; private $pwd; /*ctrl var*/ public $AccessGranted = false; function __construct($username,$password){ /