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Showing results for tags 'mysqli-prepare'.
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Hy I'm trying to execute this query $stmt = $mysqli->prepare("SELECT streamer,content,provider FROM evento,canali WHERE canali.id=evento.idcanale AND evento.titolo LIKE '%?%' OR evento.sottotitolo LIKE '%?%' AND evento.datainizio=2013-02-21;"); $stmt->bind_param('ss',$tok,$tok); $stmt->execute(); $stmt->close(); but I get this error Warning: mysqli_stmt::bind_param(): Number of variables doesn't match number of parameters in prepared statement. but to me it seems like the number are the same, you can see how I prepared the statement with 2 arguments to define, and then I passe 2 arguments to add_param, what I'm getting wrong? some has some ideas? thanks daniele New php-forum User Posts: 2 Joined: Fri May 03, 2013 6:48 pm
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I tried to use FIND_IN_SET() with prepared statement, but did work, do not return any result, or even errors if(escape($_POST['jobCategory']) != "all-categories" && escape($_POST['countryId']) == "all-countries" && escape($_POST['careerLevel']) == "all-career-levels"): $the_array = [77,181]; $job_id_imploded = implode(',',$the_array); $query = mysqli_prepare($dbConnection,"SELECT jobs.id, jobs.job_title, jobs.country_id, employers.employer_name FROM jobs LEFT JOIN employers ON jobs.employer_id = employers.employer_id WHERE job_status = ? AND FIND_IN_SET('id',?)"); mysqli_stmt_bind_param($query,'si',$job_status,$job_id_imploded); endif; mysqli_stmt_execute($query); mysqli_stmt_bind_result($query,$job_id,$job_title,$countryId,$employer_name); while(mysqli_stmt_fetch($query)){ ?> <div class="job-title"> <a href="job_post.php?job_id=<?php echo htmlspecialchars($job_id) ?>" class="job-title-link"><?php echo htmlspecialchars($job_title); ?></a> </div> <?php } // End While ?>
