Search the Community
Showing results for tags 'prepared statement'.
-
Hi All, I am adding a button that will delete many things in the system which all have the same $job_id There is going to be 10 or so tables that need to delete * where $job_id = ? Is there a better way to do this rather than just iterating my code 10 times. I was looking at just joining but isnt one benefit of prepared statements that they can be used again and again to increase speed.
-
HI All Not sure how best to describe what i am trying to do so here goes. People use my system to order food from a menu - this is not for a restaurant where you order one starter main and desert, they will be ordering 00's of meals. The table layout is as follows: ssm_menu menu_id | menu_name | menu_price ssm_menu_connection menu_id | menu_item_id | surrogate_id ssm_menu_items menu_item_id | menu_item_name | menu_item_category ssm_menu_order job_id | menu_id | menu_item_id | menu_item_qty Menu items can appear on more than one menu. If a user orders 100 of menu_item_id 1, 2 & 3 these appear on menu_id 1& 2 Menu_id 1 only contains item_id 1, 2, 3 Menu_id 2 contains menu_item 1,2,3,4,5,6,7,8,910 when querying the database, i would like the menu_id to comeback as 1 as all of the items appear on this menu and out of all available menu items the higher percentage have been picked from this menu than menu_id 2. So on menu_id 1 = 100% of the available menu_item_id have been given a menu_item_qty but on menu_id 2, only 30% of the available menu_item_id have been selected. the sql that i have started with is the following: select menu_name from ssm_menu a inner join ssm_menu_connection b on a.menu_id = b.menu_id inner join ssm_menu_items c on b.menu_item_id = c.menu_item_id inner join ssm_menu_order d on c.menu_item_id = d.menu_item_id where job_id = 27 The result of this is: Menu One, Menu One, Menu One, Menu Two, Menu Two, Menu Two I hope this is enough information to shed some light on what i am trying to achieve and appreciate any feedback in advance. Kind Regards Adam
-
I tried to use FIND_IN_SET() with prepared statement, but did work, do not return any result, or even errors if(escape($_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 ?>
- 8 replies
-
- php
- prepared statement
-
(and 1 more)
Tagged with: