Sign Up

Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.

Have an account? Sign In

Have an account? Sign In Now

Sign In

Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.

Sign Up Here

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

The Archive Base

The Archive Base Logo The Archive Base Logo

The Archive Base Navigation

  • SEARCH
  • Home
  • About Us
  • Blog
  • Contact Us
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Add group
  • Groups page
  • Feed
  • User Profile
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Buy Points
  • Users
  • Help
  • Buy Theme
  • SEARCH
Home/ Questions/Q 7659541
In Process

The Archive Base Latest Questions

Editorial Team
  • 0
Editorial Team
Asked: May 31, 20262026-05-31T13:20:47+00:00 2026-05-31T13:20:47+00:00

This is my code: $q=mysql_query(SELECT * FROM `table1` WHERE name like ‘%$searchText%’); while($e=mysql_fetch_assoc($q)) //$output[]=$e;

  • 0

This is my code:

$q=mysql_query("SELECT * FROM `table1` WHERE name like '%$searchText%'");

      while($e=mysql_fetch_assoc($q))

              //$output[]=$e;
              //echo $e['NAME'];
              {
              $name = $e['NAME'];
              $brand = $e['BRAND'];
              $category = $e['CATEGORY'];
              $query = "INSERT INTO table2 (brand, name, category) VALUES ('$brand', '$name', '$category')";
              $result = mysql_query($query) or die("Unable to insert because : " . mysql_error()); 
              }

Since in “BRAND”, there may be some data like “First’s Choice”.
In this case, I cannot insert to database due to error.
How can I insert data that contain single quotes?
Thx

  • 1 1 Answer
  • 0 Views
  • 0 Followers
  • 0
Share
  • Facebook
  • Report

Leave an answer
Cancel reply

You must login to add an answer.

Forgot Password?

Need An Account, Sign Up Here

1 Answer

  • Voted
  • Oldest
  • Recent
  • Random
  1. Editorial Team
    Editorial Team
    2026-05-31T13:20:48+00:00Added an answer on May 31, 2026 at 1:20 pm

    you need to use mysql_real_escape_string on the value, which you should be doing anyway. That should properly escape your value for insertion.

    $name = mysql_real_escape_string($e['NAME']);
    $brand = mysql_real_escape_string($e['BRAND']);
    $category = mysql_real_escape_string($e['CATEGORY']);
    $query = "INSERT INTO table2 (brand, name, category) VALUES ('$brand', '$name', '$category')";
    
    • 0
    • Reply
    • Share
      Share
      • Share on Facebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
      • Report

Sidebar

Related Questions

now i use this code: $welcome_text = mysql_query(SELECT * FROM `text` WHERE `name` =
This is my code: $query = mysql_query(SELECT * FROM books ORDER BY id) or
I have this code for print a multi columns table from mysql $k=<table width='100%'
I have this code for print a multi columns table from mysql $k=<table width='100%'
I have the following code in my C++ program: query_state = mysql_query(connection, select *
I have this code - which is trying to get variables from the URL
Im using this code for mysql connection $con = mysql_connect(localhost:/var/lib/mysql/mysql.sock, abc , xyz); if
For eg, I have this code for my PHP calendar events application. /**MySQL code
I have this mySQL code that connects to my server. It connects just fine:
I have this code $con = mysql_connect(localhost,root,); if (!$con) { die('Could not connect: '

Explore

  • Home
  • Add group
  • Groups page
  • Communities
  • Questions
    • New Questions
    • Trending Questions
    • Must read Questions
    • Hot Questions
  • Polls
  • Tags
  • Badges
  • Users
  • Help
  • SEARCH

Footer

© 2021 The Archive Base. All Rights Reserved
With Love by The Archive Base

Insert/edit link

Enter the destination URL

Or link to existing content

    No search term specified. Showing recent items. Search or use up and down arrow keys to select an item.