mysqli

问题 I am getting pretty confused trying to switch from MySQL to MySQLi, and I'm just trying a very basic example to see if I can get it to work. Please forgive the poor practice of using .post instead of .ajax and not returning JSON for now, I'm just using some old code to see if I can get queries to work. Currently, on index.php, I create a new connection: $connection = new mysqli($host, $username, $password, $database); if (mysqli_connect_errno()) die(mysqli_connect_error()); Then, using jquery

问题 I have a form and I am using jQuery validate plugin. I am also using remote to check if the username and email are available or not but I always get false back from the script. here is the jquery validation part: $(document).ready(function () { $('#registration').validate({ rules: { email: { email: true, remote:{ url: "./php/checkemail.php", type: "POST", } }, username: { minlength: 4, maxlength: 16, remote: { url: "./php/checkusername.php", type: "POST", } }, messages: { email: { email:

问题 I have an array, Array ( [0] => Array ( [userTag] => All [fbId] => 10210118553469338 [fName] => Danish [lName] => Aftab [imageUrl] => https://scontent.xx.fbcdn.net/v/t1.0-1/p50x50/22491765_10210410024475931_8589925114603818114_n.jpg?oh=7fa6266e7948ef2d218076857972f7e0 [subsType] => gold [user_visible] => 0 [distance] => 0.01 [advising] => 0 [avgRate] => 4 [totalReview] => 2 [us_seeker_type] => new [price] => 70 ) [1] => Array ( [userTag] => All [fbId] => 10210118553469338 [fName] => Danish

问题 When I use mysqli_multi_query() with a mass INSERT query, then after the query has run do another query using mysqli_query('SELECT...'), I get "commands out of sync" errors. Does anyone know why this happens? 回答1: You've to call mysqli_use_result (or mysqli_store_result) after a mysqli_multi_query() call. Lower level documentation about this error: http://dev.mysql.com/doc/refman/5.1/en/commands-out-of-sync.html 回答2: Use mysqli_next_result($db) , after freeing the result set if any . This

问题 When I use mysqli_multi_query() with a mass INSERT query, then after the query has run do another query using mysqli_query('SELECT...'), I get "commands out of sync" errors. Does anyone know why this happens? 回答1: You've to call mysqli_use_result (or mysqli_store_result) after a mysqli_multi_query() call. Lower level documentation about this error: http://dev.mysql.com/doc/refman/5.1/en/commands-out-of-sync.html 回答2: Use mysqli_next_result($db) , after freeing the result set if any . This

问题 I have a database I'm trying to display for edit through PHP. On the main page I want to show only the first 15 rows of the table then have the user click to generate more table rows. <?php include('../config.php'); $result = $mysqli->query("SELECT uid, title, description, tblFacilityHrsDateTimes.* FROM tblFacilityHrs LEFT JOIN tblFacilityHrsDateTimes ON tblFacilityHrs.uid = tblFacilityHrsDateTimes.owner_uid ORDER BY tblFacilityHrs.title") or die($mysqli->error); while($row =$result->fetch

问题 I have a database I'm trying to display for edit through PHP. On the main page I want to show only the first 15 rows of the table then have the user click to generate more table rows. <?php include('../config.php'); $result = $mysqli->query("SELECT uid, title, description, tblFacilityHrsDateTimes.* FROM tblFacilityHrs LEFT JOIN tblFacilityHrsDateTimes ON tblFacilityHrs.uid = tblFacilityHrsDateTimes.owner_uid ORDER BY tblFacilityHrs.title") or die($mysqli->error); while($row =$result->fetch

问题 I have a 11 tables email1,email2,email3,email4,email5,email6,email7,email8,email9,email10,email11 and same column name Contact_Email with different email address <?php $con = mysql_connect("localhost","root",""); $db = mysql_select_db("email-db",$con); $sql = "SELECT Contact_Email FROM email1,email2,email3,email4"; $result = mysql_query($sql); while($row = mysql_fetch_array($result)) { ?> <tr> <td><? echo $row['Contact_Email']; ?></td> <td><? echo '<br>'; ?></td> </tr> <? } ?> actually I want

问题 I'm making a login page. The connection to mysqli keeps showing me this message: Fatal error: Call to undefined function new_mysqli() in C:\wamp\www\work\conn.php on line 8 conn.php : <?php $dbhost = "loaclhost"; $dbuser = "root"; $dbpass = ""; $dbname = "users"; $conn = new mysqli($dbhost, $dbuser, $dbpass, $dbname); ?> 回答1: change $dbhost = "loaclhost"; to $dbhost = "localhost"; 来源： https://stackoverflow.com/questions/35007100/fatal-error-call-to-undefined-function-new-mysqli-in-c-wamp-www

问题 I have the following SQL to delete duplicate values form a table, DELETE p1 FROM `ProgramsList` p1, `ProgramsList` p2 WHERE p1.CustId = p2.CustId AND p1.CustId = 1 AND p1.`Id`>p2.`Id` AND p1.`ProgramName` = p2.`ProgramName`; Id is auto incremental for a given CustId ProgramName must be unique (currently it is not) The above SQL takes about 4 to 5 hours to complete with about 1,000,000 records Could anyone suggest a quicker way of deleting duplicates from a table? 回答1: First, You might try