Php query function (how to obtain the fields of the corresponding table through the query conditions)

Php query function (how to obtain the fields of the corresponding table through the query conditions)

 "; // Connect to the Database mysql_select_db (" test ") or die (mysql_error (); // echo" Connected to Database "; // query data, use the table to Display @ $ checked =$ _ POST ['checked']; @ $ result = mysql_query ("select * from books", $ db); echo"
 
 
  
  
\ N "; echo"
  
  "; Echo" 
   \ N "; // Loop traversal while ($ myrow = mysql_fetch_row ($ result) {printf (" 
   ", $ Myrow [0], $ myrow [1], $ myrow [2], $ myrow [3]);} echo" 
   
 
   
 
    
Isbn
 
    
Author
 
    
Title
 
    
Price
 
    
% S
 
    
% S
 
    
% S
 
    
% S
 
  
 
 
\ N ";?>
 Untitled Document

Reply to discussion (solution)

Please describe your problem

Result = mysql_query ("select * from books where checked = '$ checked '");

For example, enter the isbn number to be queried, and then confirm, and output the matching query results below.

Result = mysql_query ("select * from books where checked = '$ checked'"); I tried this and the following error occurred.
Another question is how to display the results of a table after the query.

... Where field = '$ checked' // replace the field with your actual field

If (isset ($ _ POST ['checked']) {
@ $ Result = mysql_query ("select * from books", $ db );
Echo"

\ N ";Echo" ";Echo" \ N ";// Loop traversalWhile ($ myrow = mysql_fetch_row ($ result )){Printf (" ", $ Myrow [0], $ myrow [1], $ myrow [2], $ myrow [3]);}Echo"

Isbn	Author	Title	Price
% S	% S	% S	% S

\ N ";
}

Thank you, jordan102. how can I use the query conditions to output tables that meet the query conditions on the same page?
This is my table.

I tried jordan102 code and I do not know how to add if (isset ($ _ POST ['checked']) {......} the location of this code is incorrect.
I pressed the query button but did not respond. The code is as follows:

 "; // Connect to the Database mysql_select_db (" test ") or die (mysql_error (); // echo" Connected to Database "; // query data, use the table to Display @ $ checked =$ _ POST ['checked']; // @ $ result = mysql_query ("select * from books", $ db ); // @ $ result = mysql_query ("select * from books where isbn = '$ checked'"); if (isset ($ _ POST ['checked']) {@ $ result = mysql_query ("select * from books", $ db); echo"
 
 
  
  
\ N "; echo"
  
  "; Echo" 
   \ N "; // Loop traversal while ($ myrow = mysql_fetch_row ($ result) {printf (" 
   ", $ Myrow [0], $ myrow [1], $ myrow [2], $ myrow [3]);} echo" 
   
 
   
 
    
Isbn
 
    
Author
 
    
Title
 
    
Price
 
    
% S
 
    
% S
 
    
% S
 
    
% S
 
  
 
 
\ N ";}?>
 Untitled Document

@ $ Result = mysql_query ("select * from books where isbn = '$ checked'", $ db );

@ Jordan102 I tried @ $ result = mysql_query ("select * from books where isbn = '$ checked'", $ db). but I still cannot click the button.

This sentenceI tried it, but when I input the isbn of books to the input box, I click query. I still cannot find all the corresponding content of isbn = 1.

Only a portion of the data is available. check whether the data is different.

$ Result = mysql_query ("select * from books", $ db );
If none of the conditions are added, is it a condition query?

No data table is displayed after I run it. The result is as follows:
Will there be no matching for those fields?

Sorry! Forgot to add.

If no matching record exists, the natural list is empty.

@ $ Result = mysql_query ("select * from books where isbn = '$ checked'", $ db );
Where is the value assigned ???

Hey, I marked it red on #18 and you still copied the error.

// Obtain the connection
$ Db = mysql_connect ("localhost", "root", "root") or die (mysql_error ());
// Echo "Connected to MySQL
";
// Connect to the database
Mysql_select_db ("test") or die (mysql_error ());
// Echo "Connected to Database ";

// Query the data and display it in a table
// @ $ Checked = $ _ POST ['checked'];
// @ $ Result = mysql_query ("select * from books", $ db );
// @ $ Result = mysql_query ("select * from books where isbn = '$ checked '");
If (isset ($ _ POST ['checked']) {
$ Checked = $ _ POST ['checked'];
@ $ Result = mysql_query ("select * from books where isbn = '$ checked'", $ db );
Echo"

\ N ";Echo" ";Echo" \ N ";// Loop traversalWhile ($ myrow = mysql_fetch_row ($ result )){Printf (" ", $ Myrow [0], $ myrow [1], $ myrow [2], $ myrow [3]);}Echo"

Isbn	Author	Title	Price
% S	% S	% S	% S

\ N ";
}
?>



Untitled Document

Reply to qxhaidao and jordan102: After one afternoon, the qxhaidao code runs successfully, and jordan only displays the header. I will add two points. thank you!