如何解决未定义索引：在php中显示数据库检索值时的注意事项

Question

I'm getting an undefined index notice, but can't figure out how to fix it.

Below is the code which is generating the error:

<?php

include 'dbconfig.php';

$id;

if(isset($_POST['Tid']))
{
  $id=$_POST['To_ID'];
  echo "selected value:".$id;
  $query="select 'ID','Name' from  where ID='$id'";
  $result=sqlsrv_query($conn,$query);
  $a=sqlsrv_num_rows($result);
  echo $a;

  if(!$result)
  {
    echo "Query execution failed"; 
  }   
  else {
    while($row = sqlsrv_fetch_array($result))
    {
       print_r($row['ID'].$row['Name']);
    }
  } 
} 
else {
  echo "Post variable not set";
}

I am getting error on the line here-

print_r($row['ID'].$row['Name']);

And also i am verified database columns the columns ID and Name is present in the database and the select query is also executing well when i tried seperately.

Even though the filed name ID and Name exists i am getting Undefined Index ID and Name Error.

Please help me to resolve this error.

Thanks in advance.

Answer 1

You are using the wrong kind of quotes and you are missing the table name:

$query="select 'ID','Name' from  where ID='$id'";

Should be:

$query="select ID,Name from table_name where ID='$id'";

Or, if the column names need to be quoted you need block-quotes in sql server:

$query="select [ID],[Name] from [table_name] where [ID]='$id'";

Also note that you have an sql injection problem. Ideally, you should use a prepared statement, but if the value of the ID is an integer, you can also do:

$id = (int) $_POST['To_ID'];
$query = "select ID,Name from table_name where ID='$id'";

Answer 2

try this:

$query="select ID,Name from where ID='$id'";

$result=sqlsrv_query($conn,$query);


$a=sqlsrv_num_rows($result);
echo $a;

if(!$result)
{
echo "Query execution failed"; 
}    else {

for($i = 0; $i < $a; $i++)
{

print_r($row[$i]['ID'].$row[$i]['Name'].'<br/>');

}
} 
} else {
echo "Post variable not set";
}