mysqli

问题 The following code is throwing the mysterious Warnings. I can't understand what they mean. What do these errors indicate and how to eradicate them? require "conn.php"; $q = mysqli_stmt_init($dbconn); $query = "SELECT users.userid FROM users WHERE users.email = ? "; mysqli_stmt_prepare($q, $query); mysqli_stmt_bind_param($q, "s", $email); mysqli_stmt_execute($q); $result = mysqli_stmt_get_result($q); if (mysqli_num_rows($result) == 0) { $q = mysqli_stmt_init($dbconn); $query = "INSERT INTO

问题 I'm using PHP and mysqli prepared statements. Is there a compelling reason to manually check for errors when executing mysqli_stmt_prepare() ? To be more specific I am not asking about the final result just the prepare statement line. $sql = "SELECT * FROM `users`;"; $stmt = mysqli_stmt_init($db); mysqli_stmt_prepare($stmt, $sql); // How should I check for error in here mysqli_stmt_execute($stmt); $result = mysqli_stmt_get_result($stmt); PHP manual puts this and only this line in an if

问题 I'm using PHP and mysqli prepared statements. Is there a compelling reason to manually check for errors when executing mysqli_stmt_prepare() ? To be more specific I am not asking about the final result just the prepare statement line. $sql = "SELECT * FROM `users`;"; $stmt = mysqli_stmt_init($db); mysqli_stmt_prepare($stmt, $sql); // How should I check for error in here mysqli_stmt_execute($stmt); $result = mysqli_stmt_get_result($stmt); PHP manual puts this and only this line in an if

问题 I understand that we have to pass references to mysqli_stmt_bind_param . I am doing the following $ref = refValues($data); function refValues($arr){ $refs = array(); foreach($arr as $key => $value) $refs[$key] = &$arr[$key]; var_dump(implode(",", $refs)); return $refs; return $arr; } I am having all of my values in an array. I am using the above function to get the references. Got the above answer from SO My PHP version is 5.6 I am binding the params in the following way. mysqli_stmt_bind

问题 Alright so this bugs the crap out of me, and I can't seem to find anything in the PHP documentation, nor anywhere in the Google resultosphere, so maybe someone can help here. I'm trying to make a simple request to a database to return a varchar(30) . Code: $qryID = "select idPhotos from housesphotos where idHouse = ?"; //Prepare the query $stmt = mysqli_prepare($link, $qryID); $x = 106; //Bind the login parameter to the statement if(mysqli_stmt_bind_param($stmt, "i", $x)){ if(mysqli_stmt

问题 Closed . This question needs details or clarity. It is not currently accepting answers. Want to improve this question? Add details and clarify the problem by editing this post. Closed yesterday . I am working with WordPress and it is showing this error: WordPress is also running slowly but I don't know why. I am using WAMP on Windows 8. 来源： https://stackoverflow.com/questions/62101777/warning-mysqli-query-hy000-06

问题 I am using mysqli bind_param() when I'm inserting stuff to my DB. When I tried to insert a phone number as "0737327373" the value got inserted as "737327373" without the leading "0". But when I changed bind_param("i", $phoneNumber) to bind_param("s", $phoneNumber) instead the value got inserted correctly as "0737327373". So should I use "s" instead of "i" when I'm inserting numbers with a leading 0 in it? Also, I have a value that could be anywhere between (0-5) and (0-1) depending on what

问题 I'm working on streamlining a bit our db helpers and utilities and I see that each of our functions such as for example findAllUsers(){....} or findCustomerById($id) {...} have their own connection details for example : function findAllUsers() { $srv = 'xx.xx.xx.xx'; $usr = 'username'; $pwd = 'password'; $db = 'database'; $port = 3306; $con = new mysqli($srv, $usr, $pwd, $db, $port); if ($con->connect_error) { die("Connection to DB failed: " . $con->connect_error); } else { sql = "SELECT *

问题 I'm working on streamlining a bit our db helpers and utilities and I see that each of our functions such as for example findAllUsers(){....} or findCustomerById($id) {...} have their own connection details for example : function findAllUsers() { $srv = 'xx.xx.xx.xx'; $usr = 'username'; $pwd = 'password'; $db = 'database'; $port = 3306; $con = new mysqli($srv, $usr, $pwd, $db, $port); if ($con->connect_error) { die("Connection to DB failed: " . $con->connect_error); } else { sql = "SELECT *

问题 PHP login system works on local but doesn't work on Hostgator PHP 7.1. My local is PHP 7.2 I've built out a fully working portal on my local machine. I can CRUD on my local machine. As soon as I put it on the server online, the login system doesn't work. I still can register new users as the user info populates in the DB, so its not a DB config issue. I am getting these errors: Warning: mysqli_stmt_bind_param(): Number of variables doesn't match number of parameters in prepared statement in .