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khunjack

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  1. mac_gyver,Jacques1 Thanks to you both for your replies and help. Very appreciated (What a great board). Made me find a routine. I just checked and got - MySQL is installed. - MySQLi is installed. - libmysqlclient driver is being used. - PDO MySQLnd is enabled. Seems I will be ok using PDO I used :
  2. Jacques1, Thank you very much for your fast reply. I understand the external server management company I use, (not my host/server company) explanation is misleading. Do you see a way for me to see, if it is disabled and to enable the native driver ? Thanks a lot. Jack PS: I wasn't sure if for new dev. I would use mysqli or PDO. I will go for PDO. PS: I'm a little bit reluctant to tell them. They just tried to upgrade to Easyapache 4 which didn't work with mod_php and all sites on this server where not accessible for nearly 24 hrs.
  3. Hi, Using a new external script and I'm getting the following error: Fatal error: Call to undefined function mysqli_stmt_get_result() in /home/.... According to the script developer, an host issue. I asked my server management company and they came back with this answer: Note: o currently I need to stick with mod_php and php 5.4.45 o According to phpinfo: mysql, mysqli, mysqlnd is installed. If it is correct that "mysqli_stmt_get_result" is only avaialble till php 5.3? Any suggestion how to tackle this issue? Thanks a lot in advance Jack
  4. Mchl, Thank you for your reply. $aRow = $this->getPicture($sPicId); . $src = @imagecreatefromstring ($aRow["Bin Data"]); $image = imagecreatetruecolor (...); imagecopyresampled (...); . . $rotate = imagerotate($image, $degrees, 0) ; imagejpeg($rotate, $imagename2) ; now I have the rotated pic in '$imagename2' in the server root Up to now could not manage to store the rotated image back into mysql. Tried a few things, honestly I don't have a glue, e.g. stored the $rotate, $imagename2 in the blob ["Bin Data"] field... etc... Thank you Jack
  5. PS: The image is on the server root .... the issue is to store it back into mysql Thanks
  6. Dears, I have images in mysql (blob). Some of the images need to be rotated (90 degrees). Not sure how to do it. Do I have to physical write the image? (I did that and was using 'imagerotate' -> image rotated, but couldn't manage to store the rotated images back into mysql.) Can the blob be rotated?? Any hints/suggestions are very welcome. Thanks a lot Jack
  7. I'm puzzled .... made it very simple SELECT COUNT(DISTINCT accounts.account_id) FROM kj_accounts accounts JOIN kj_stats_sys_notification sysnotify ON (accounts.account_id=sysnotify.account_id) and it seems to work fine I tried that already earlier but without the 'DISTINCT ' It helped big time pointing that out. Guys was great that you responded so fast to my help request. Thanks a lot.
  8. BTW: I need that count for paging (total record #). to get the actual entries in table 2 (counting how many sysnotification that id got) seems to work SELECT COUNT(sysnotify.sender_site) AS mail_cnt, accounts.account_id, account_screenname FROM kj_accounts accounts RIGHT JOIN kj_stats_sys_notification sysnotify ON (accounts.account_id=sysnotify.account_id) GROUP BY accounts.account_id ORDER BY mail_cnt desc, created_dt desc LIMIT 0, 10 I need the count for that, sure I think the work around would be using above without LIMIT and doing an count($aResult) wouldn't fit into all my other routines .... drives me nuts
  9. Thanks a lot again .... all make so much sense, when I look at it ...but Result (for both): 1 used: SELECT COUNT(DISTINCT accounts.account_id) AS number_of_rows FROM kj_accounts accounts left JOIN kj_stats_sys_notification sysnotify ON (accounts.account_id=sysnotify.account_id) WHERE sysnotify.account_id is not null GROUP BY accounts.account_id and SELECT COUNT(DISTINCT accounts.account_id) AS number_of_rows FROM kj_accounts accounts JOIN kj_stats_sys_notification sysnotify ON (accounts.account_id=sysnotify.account_id) GROUP BY accounts.account_id ..thanks ..thanks for any other ideas ... so much appreciated
  10. Thanks a lot toplat and colldude832 I tried the first eaxample from toplay as SELECT COUNT(accounts.account_id) AS number_of_rows FROM kj_accounts accounts left JOIN kj_stats_sys_notification sysnotify ON (accounts.account_id=sysnotify.account_id) WHERE sysnotify.account_id is not null Result: 261 (I'm expecting 126) table1 : accounts (158 records) table2: sysnotify (261) tried cooldude832 suggestion as Select IF(COUNT(sysnotify.account_id) >= '6') accounts.account_id AS account_id from kj_stats_sys_notification sysnotify, kj_accounts accounts that brings an syntax error: 1064 ** You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') accounts.account_id AS account_id from kj_stats_sys_notification sysnotify, kj' at line 1 Thanks so far a lot for your help
  11. Dears, as longer I'm trying as more I get confused. I have 2 tables table 1 has records with 'account id' unique (only once) table 2 has 'account it' multiple times (one to many relationship) I want just counting the recs in table 1 which having one or many records in table 2 The wished result is: count only one when 'account id' exist in table 2 ( even when table 2 has e.g. 10 records with the 'account id' from table 1) Dazed and confused ..... Any help/hint is highly appreciated - Thanks a lot in advance
  12. Barand, fantastic!!! Thank you very much for your straight forward solution .. big time appreciated. Jack
  13. Hi, I'll have a session array with growing numbers of id's. array = (2222,4444,32323,...} The records with these id's need to be excluded ... e.. ' .... id!=2222 and id!= 4444 ' Sure could build this querey via a loop. Question: Is there a better/shorter way in the WHERE Clause ??? ??? WHERE id!={array} ... ??? Thanks for every feedback Jack
  14. Hi, I try to display Thai on my page. e.g (แจ็ค) but in the html .. it looks like the following ( 3649; 3592;& #3655;& #3588;) (I put blanks in it here..to show) is there any other way ...that I could type that like (แจ็ค) in the html. What do I need to do that it will not displayed like '????????' for the user. Thanks a lot - a little bit confused Kindest regards Jack
  15. What do I do wrong. I use charset=utf-8 or charset=tis-620 or windows-874. I copy the Thai text into the page ... can see it perfectly - store the page. Call via a Browser ... all "???????????????????". Open it up with Dreamweaver ... now also "???????????????????". What do I do wrong .. what do I need to do to create a page in Thai LAnguage Thanks a lot for your help .. Highly appreciated. Jack (แจ็ค)
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