在PHP中回显变量的最佳方法[关闭]

Question

From what I've read, it isn't possible to echo more PHP code within an echo so I'm after a solution to the following:

Basically, if the 'closed_state' column in my DB equals 'yes', display a link with a variable on the end else display a different link with a variable on the end.

Here's my code:

<?php 

$result1    = mysqli_query($con,"SELECT closed_state FROM tbl_company WHERE company_id='$company_id'") or die(mysql_error());
$result2    = mysqli_query($con,"SELECT company_id FROM tbl_company WHERE company_id='$company_id'") or die(mysql_error());
$result3    = mysqli_query($con,"SELECT company_id FROM tbl_company WHERE company_id='$company_id'") or die(mysql_error());

while($row = mysqli_fetch_array($result1))
{
    if ($row['closed_state'] == "yes")
    {
        echo "<a href="customers_open.php?company_id=<?php while($row = mysqli_fetch_array($result2)) echo "{$row['company_id']}"; ?>">Reopen account</a>";
    } else {  
        echo "<a href="customers_close.php?company_id=<?php while($row = mysqli_fetch_array($result3)) echo "{$row['company_id']}"; ?>">Close account</a>";
    }
}   

?>

Obviously it doesn't work so what's the best way to achieve this?

Answer 1

I don't think you need 3 queries. One should be fine selecting both values. Queries 2 & 3 are identical anyway.

$result1    = mysqli_query($con,"SELECT company_id, closed_state FROM tbl_company WHERE company_id='$company_id'") or die(mysql_error());
while($row = mysqli_fetch_array($result1)){
    if ($row['closed_state'] == "yes"){
        echo '<a href="customers_open.php?company_id='.$row['company_id'].'">Reopen account</a>';
    }else{  
        echo '<a href="customers_close.php?company_id='.$row['company_id'].'">Close account</a>';
    }
}