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Mysql Job figure out why lastlogin is no longer working / showing in DB

$10-30 USD

Fullført
Lagt ut omtrent 4 år siden

$10-30 USD

Betalt ved levering
Small $10 Job to figure out why --- mysqli_query($linkID, "UPDATE $database SET lastLogIn='$currentTime' WHERE id = '$tempId' LIMIT 1"); is no longer posting to DB $10 Job
Prosjekt-ID: 24420797

Om prosjektet

12 forslag
Eksternt prosjekt
Aktiv 4 år siden

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Tildelt til:
Brukeravatar
Hello, I am an expert in Backend Development. I can help you to fix the MySQL error in your Backend PHP Code. Feel free to contact me right now!
$10 USD om 1 dag
5,0 (12 omtaler)
3,4
3,4
12 frilansere byr i gjennomsnitt $14 USD for denne jobben
Brukeravatar
I am experienced PHP developer and web designer with the depth knowledge related to MySQL,Codeigniter,Laravel,WordPress,MSSQL,JSON,JAVASCRIPT,AJAX,XML,jQuery,CSS with the integration of web technologies. I have worked as PHP & Web developer in industry and now I am full time freelancer and having the experience of huge and live projects. I am finding one startup like you so i can provide my skills to IT market and I can give you 100% surity for quality work in minimum time. I can work at anytime with the great satisfactions,quality output and result in your hand by fulfilling your requirements. Waiting for your positive response , so we can start the work soon ...
$10 USD om 1 dag
5,0 (539 omtaler)
7,8
7,8
Brukeravatar
I have a few questions regarding the specifications of the project. Could we please have a chat?
$10 USD om 14 dager
4,7 (67 omtaler)
5,9
5,9
Brukeravatar
Hi there UPDATE $database SET lastLogIn='$currentTime' WHERE id = '$tempId' LIMIT 1 it should'nt be $database ! please award me your project, ill make it work within minutes Regards Zaky.
$10 USD om 1 dag
4,9 (15 omtaler)
4,4
4,4
Brukeravatar
Hello, I am web developer with 7 years of experience using php, mysql i can fix your lastlogin problem Thanks Jaffer I.
$15 USD om 1 dag
5,0 (22 omtaler)
3,9
3,9
Brukeravatar
Hello Just read your project. Your query is wrong. mysqli_query($linkID, "UPDATE". $database ."SET lastLogIn='".$currentTime."' WHERE id = '".$tempId."' LIMIT 1"); this is the correct query. Please message me.
$10 USD om 1 dag
5,0 (11 omtaler)
3,5
3,5
Brukeravatar
i can figure out and fix mysqli query update query not working. i have knowledge of mysqli database and queries
$15 USD om 1 dag
5,0 (3 omtaler)
2,5
2,5
Brukeravatar
Hi! UPDATE commands is usually followed by "[databasename]"."[tablename]". Maybe your $database variable doesn't have the table name in it.
$10 USD om 1 dag
0,0 (0 omtaler)
0,0
0,0
Brukeravatar
How to do work online in tha job
$25 USD om 1 dag
0,0 (0 omtaler)
0,0
0,0

Om klienten

UNITED STATESs flagg
Perris, United States
5,0
41
Betalingsmetode bekreftet
Medlem siden apr. 15, 2012

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