How to hide the value of the drop-down box-php Tutorial

How to hide the value of the drop-down box?

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If there is a service class record in the database

So hide it when you submit the form next time. ServiceThis

How to implement it?

Reply to discussion (solution)

 Select
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 Select
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How smart are you? I never thought of the in_array () function.

But why can't I set the value?

Suppose my data TABLE is a TABLE with three pieces of information.
Id uid name
1, 2, 3
2 8 Li Si
3 9 Wang 'er

$ Result = mysql_query ("SELECT * FROM uid", TABLE );
$ Row = mysql_num_rows ($ result );
Incorrect output $ row

$ Result = mysql_query ("SELECT * FROM uid", TABLE );
Obvious problems with the from uid


Change
$ Uid = xxx; // The uid to be queried
$ Result = mysql_query ("SELECT * from table where uid = '". $ uid. "'") or die (mysql_error ());
Try

$ Result = mysql_query ("SELECT * FROM uid", TABLE );
Obvious problems with the from uid


Change
$ Uid = xxx; // The uid to be queried
$ Result = mysql_query ("SELECT * from table where uid = '". $ uid. "'") or die (mysql_error ());
Try

I didn't want to query a piece of information, didn't you say it on the second floor upstairs? I want to use this method to get all the uid data output results: 2, 8, 9 arrays of this type
Use in_array () to determine the drop-down list.

That's simpler.

$query = mysql_query("select uid from TABLE") or die(mysql_error());$result = array();while($thread=mysql_fetch_assoc($query)){    $result[] = $thread['uid'];}

$ Result contains the existing uid.

That's simpler.

$query = mysql_query("select uid from TABLE") or die(mysql_error());$result = array();while($thread=mysql_fetch_assoc($query)){    $result[] = $thread['uid'];}

$ Result contains the existing uid.

Output is Array
Solution

That's simpler.

$query = mysql_query("select uid from TABLE") or die(mysql_error());$result = array();while($thread=mysql_fetch_assoc($query)){    $result[] = $thread['uid'];}

$ Result contains the existing uid.

Is there a way to use the while loop or while external output?

Why not bring $ result into $ exists?

Why not bring $ result into $ exists?

No.

That's simpler.

$query = mysql_query("select uid from TABLE") or die(mysql_error());$result = array();while($thread=mysql_fetch_assoc($query)){    $result[] = $thread['uid'];}

$ Result contains the existing uid.

Output is Array
Solution

Array output is of course Array

So you should understand.

$ Query = mysql_query ("select uid from TABLE") or die (mysql_error (); $ result = array (); while ($ thread = mysql_fetch_assoc ($ query )) {$ result [] = $ thread ['uid'] ;}?>Select
  Service
  
  Manufacturing
  
  Advertising
  

That's simpler.

$query = mysql_query("select uid from TABLE") or die(mysql_error());$result = array();while($thread=mysql_fetch_assoc($query)){    $result[] = $thread['uid'];}

$ Result contains the existing uid.

Output is Array
Solution

Array output is of course Array

So you should understand.

$ Query = mysql_query ("select uid from TABLE") or die (mysql_error (); $ result = array (); while ($ thread = mysql_fetch_assoc ($ query )) {$ result [] = $ thread ['uid'] ;}?>Select
  Service
  
  Manufacturing
  
  Advertising
  

Thank you.