prepared-statement

How to prepared statement for IN clause if there is mixed type array?

淺唱寂寞╮ 提交于 2019-12-24 05:05:35
问题 I am trying to write prepared statement for IN clause where array is mixed array. i am able to build array for 'call_user_func_array()' but i am not able to prepare statement. no output is showing. this is my php code: $search1 = array('pune','india','2014','mumbai','2015'); print_r($search1); echo $param = implode(",",$search1); $ids = array_flip(array_flip(explode(',', $param))); var_dump($ids); $type1 = array(); foreach ($ids as $element) { if(is_string($element)) { $type1[] ='s'; }elseif

MySql “view”, “prepared statement” and “Prepared statement needs to be re-prepared”

岁酱吖の 提交于 2019-12-24 04:54:20
问题 I've a problem with MySql, here is the details : I created a new schema/database, executed (only) this queries : create table mytable ( id varchar(50) not null, name varchar(50) null default '', primary key (id)); create view myview as select id,name from mytable; insert into mytable values ('1','aaa'); insert into mytable values ('2','bbb'); insert into mytable values ('3','ccc'); and then, if I run these queries : select * from mytable; select * from myview; prepare cmd from 'select id,name

PDO not binding placeholders

元气小坏坏 提交于 2019-12-24 04:18:46
问题 I am trying to change my log in script from mysql to PDO . For the rest of my script all seams to be going well apart from this parts and I simply cant see why. I have the below code ... $pasword=md5($_POST['password']); $email=$_POST['email']; .... $query ="SELECT id FROM guests WHERE email=':eml' AND password =':pwd' AND lead_guest=17"; // $param2=array(':eml'=>$email,':pwd'=>$pasword); $state=$dbh->prepare($query); $state->bindParam(':eml',$email); $state->bindParam(':pwd',$pasword);

Prepared statement works for INSERT but not for SELECT

北城以北 提交于 2019-12-24 04:18:11
问题 EDIT: SOLVED, SEE BELOW But still dont know what did I do to make it work :) OK, I am stuck now. I have table users : ID int PRIMARY AUTO_INCREMENT EMAIL varchar(60) NICK varchar(60) //... If I do: <?php $email = $_POST["mai"]; $nickname = $_POST["nck"]; $mysqli = new mysqli($db_host, $db_username, $db_password, $database); $prepared_statement = "INSERT INTO users VALUES(?,?,?)"; if ($stmt = $mysqli->prepare($prepared_statement)) { $id = ""; $stmt->bind_param("iss",$id,$email,$nickname);

Is there any way to use wpdb prepare statements for array implode(' OR ', $myArray)?

穿精又带淫゛_ 提交于 2019-12-24 04:01:33
问题 I'm trying to properly prepare my data for $wpdb $region = $wpdb->get_results( $wpdb->prepare( " SELECT tr.*, count(*) AS jobs_count FROM {$wpdb->terms} tr INNER JOIN {$wpdb->term_taxonomy} tm ON ( tm.term_id = tr.term_id ) INNER JOIN {$wpdb->term_relationships} tmr ON ( tmr.term_taxonomy_id = tm.term_taxonomy_id ) INNER JOIN {$wpdb->posts} p ON ( p.ID = tmr.object_id ) WHERE (tm.parent = '%d' AND tm.taxonomy = '%s' AND p.post_type = '%s' ) GROUP BY name HAVING COUNT(name) > '%d' ", 0,

Determining if no rows are returned from a prepared statment

夙愿已清 提交于 2019-12-24 03:14:49
问题 I'm implementing prepared statements on my already working mysqli queries. I'm having trouble with a line if(mysqli_num_rows($result) == 0) as it's now a string instead of a mysqli_result. if($nameAvailableStmt = mysqli_prepare($link, 'SELECT name FROM usetbl1 WHERE name=? LIMIT 1')) { mysqli_stmt_bind_param($nameAvailableStmt, "s", $_POST['dangerous']); mysqli_stmt_execute($nameAvailableStmt); mysqli_stmt_bind_result($nameAvailableStmt, $result); mysqli_stmt_fetch($nameAvailableStmt); } if

MySQLi update prepared statement not updating database

余生颓废 提交于 2019-12-24 02:45:13
问题 So I have this update statement which when I dump the $_POST variables., I get the outputs I want. $stmt = $dbConnectionW->prepare("UPDATE members SET fname='". mysqli_real_escape_string($dbConnectionW, $_POST['fname']) ."', sname='". mysqli_real_escape_string($dbConnectionW, $_POST['sname']) ."', gender='". mysqli_real_escape_string($dbConnectionW, $_POST['gender']) ."', nationality='". mysqli_real_escape_string($dbConnectionW, $_POST['nation']) ."', year='". mysqli_real_escape_string(

MySQLi update query issue [duplicate]

a 夏天 提交于 2019-12-24 01:53:52
问题 This question already has an answer here : mysqli_fetch_assoc() expects parameter / Call to a member function bind_param() errors. How to get the actual mysql error and fix it? (1 answer) Closed 7 months ago . My page sets signin_time sets it to NOW() , sets logged_in to 1 when user successfully signs into page. What I want to do is to set signout_time to NOW() when user signs out too. For this purpose I'm using following query $stmt = $db->prepare("UPDATE `ulog` SET `logged_in`=0, `signout

PDO truncates my SQL string in Prepared Statement, tries to bind parameter number 0?

坚强是说给别人听的谎言 提交于 2019-12-24 01:37:14
问题 Okay, so I've spent a couple days trolling google and finding absolutely nothing on this. Basically I'm using a prepared statement that has a couple joins and matches based on several LIKEs, an IN, and a FREETEXT. I'm building the query dynamically to support the IN condition. I think some code will help public function searchForProjWithTags(Array $dSearchArgs, $dTags) { $bSplitTags = preg_split ("/[,]/", $dTags); $bCount = count($bSplitTags); $bQuestionMarks = "?" . str_repeat (",?", $bCount

Setting Array in preparedstatement, throws a java.sql.SQLFeatureNotSupportedException

﹥>﹥吖頭↗ 提交于 2019-12-24 01:27:14
问题 I have a List of Strings that I want to set as a parameter in a preparedstatement. This question and answer here, makes it look easy: How to use an arraylist as a prepared statement parameter Well, not really easy. There is still the conversion of a List, to an SQL-Array, which I found easiest to do by creating a String[] in between. Below is my code: PreparedStatement s = con.prepareStatement("SELECT * FROM Table WHERE Country IN ? "); String[] countryArray = new String[countryListObject