mysqli

问题 SELECT * FROM datatable WHERE Name LIKE ? OR Code LIKE ? OR Date LIKE ? OR Inserter LIKE ? AND ID = '2' There is an error in php sql connection: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement in Here is the error code, I am using stmt->prepare, stmt->execute and so on in php. This works fine with just one LIKE ? (WHERE Name LIKE ? AND ID = '2'). How can I replace or solve this problem? The real code if ($stmt = $db->prepare($SearchQuery

问题 I am using the entries of a db to fill a row and a column in a table. But I cannot access the SQL returned data twice using mysqli_fetch_array() twice. This doesn't work: //Copy the result $db_res = mysqli_query( $db_link, $sql ); $db_res2=$db_res; //Top row while ($row = mysqli_fetch_array( $db_res, MYSQL_ASSOC)) { echo "<td>". $row['Title'] . "</td>"; } //leftmost column while ($row = mysqli_fetch_array( $db_res2, MYSQL_ASSOC)) { echo "<tr>"; echo "<td>". $row['Title'] . "</td>"; ..... echo

问题 I'm using MySQLi connection and I have the following code, i am trying to select all columns for a row defined by inventory id from a database table but i can't get this to work. Am i doing it the totally wrong way? I want to be able to use things like $row['general_cleanliness'] in my code following the select query. $getScheduleCondition = $db->prepare("SELECT * FROM inventories WHERE inventory_id = ? LIMIT 1",)){ $getScheduleCondition->bind_param("i", $inventory); $row =

问题 I am They try to use mysqli :: bind_param unsuccessfully . In the code , below , the function login ( ) is called with a username and password on the page login.php . Despite all efforts and guides and forums have read , continues to return the same error . I tried both with bind_param ( 's' , $ variable ) that with bind_param ( 1 , $ variable ) that with bind_param ( 'ss' , $ variable1 , $ variable2 ) and i tried query without '' "SELECT id,org_id,org_group_id,people_id FROM users WHERE

问题 I am They try to use mysqli :: bind_param unsuccessfully . In the code , below , the function login ( ) is called with a username and password on the page login.php . Despite all efforts and guides and forums have read , continues to return the same error . I tried both with bind_param ( 's' , $ variable ) that with bind_param ( 1 , $ variable ) that with bind_param ( 'ss' , $ variable1 , $ variable2 ) and i tried query without '' "SELECT id,org_id,org_group_id,people_id FROM users WHERE

问题 I need translate to "pt_br" the currently month and this code are working like a charm on phpmyadmin: SET lc_time_names = 'pt_BR'; SELECT MONTHNAME(CURDATE()) AS `Mes` FROM trabalho_v2 Someone may help me use the "SET lc_time_names = 'pt_BR';" on my php code? <?php // Show current month include 'conection.php'; $sql .= "SELECT MONTHNAME(CURDATE()) AS `Mes` FROM trabalho_v2 "; $busca = mysqli_query($conexao,$sql); $dados = mysqli_fetch_array($busca); $mes = $dados['Mes']; I tried to create a

问题 So I had this chunk of php code if($_POST['action']=='newComment') { $mysqli = new mysqli("localhost", "root", "", "nested_comment"); $new_post = $mysqli->real_escape_string($_POST['content']); $result = $mysqli->query("SELECT @myLeft := lft FROM comment WHERE lft = '1'; UPDATE comment SET rgt = rgt + 2 WHERE rgt > @myLeft; UPDATE comment SET lft = lft + 2 WHERE lft >= @myLeft; INSERT INTO comment(content, lft, rgt) VALUES('$new_post', @myLeft, @myLeft + 1);"); if($result) echo "ok"; else

问题 If I run this code, the if-clause won't return false even though the data record doesn't exist in the database myDB in table myTable . I don't know what's wrong.. // MySQLi-Connection... $result = $mysqli->query("SELECT * FROM `myDB`.`myTable` WHERE `itemtype` = 'comment' AND `itemID` = 3"); if ($result) { echo "record found!"; } else { echo "record not found!"; } The record with itemID = 3 doesn't exist, but my if-clause says that the $result returns true.. 回答1: Returns FALSE on failure. For

问题 I am trying to achieve the following: User 1: - Alert 1 Email - Alert 2 Email User 2: - Alert 1 Email - Alert 2 Email I'm trying to accomplish this with a while loop inside another while loop that is running a mysqli prepared statement, but I cannot get it to work. Code: $stmtAdd = $conn->prepare("INSERT INTO Data (row1, row2, row3, row4) VALUES ('".$row1."', '".$row2."', '".$row3."', '".$row4."')"); $stmtAdd->execute(); $stmtAdd->close(); $stmtUsers = $conn->prepare("SELECT username,

问题 I am trying to achieve the following: User 1: - Alert 1 Email - Alert 2 Email User 2: - Alert 1 Email - Alert 2 Email I'm trying to accomplish this with a while loop inside another while loop that is running a mysqli prepared statement, but I cannot get it to work. Code: $stmtAdd = $conn->prepare("INSERT INTO Data (row1, row2, row3, row4) VALUES ('".$row1."', '".$row2."', '".$row3."', '".$row4."')"); $stmtAdd->execute(); $stmtAdd->close(); $stmtUsers = $conn->prepare("SELECT username,