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Image display dependent on database call

开发者 https://www.devze.com 2022-12-16 16:05 出处:网络
$result=mysqli_fetch_array(mysqli_query($link, $query)); the IF MAYOR image is not working i guess there is a confilct with the original db call?
    $result=mysqli_fetch_array(mysqli_query($link, $query)); the IF MAYOR image is not working i guess there is a confilct with the original db call?
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this code tells the page whether or not to display some text... i want to display an alternate image to THIS ONE IF THE CONDITION IS MET... borrowing power =mayor

<div style="position:absolute;top:160px;left:535px;"><img src="images/NEI_icon_color.jpg"></div>


first detect the condition at the top of the script:

<?php
  $isMayor = false;
  ...
  if ($row['borrowingPower'] == 'mayor') {
    $isMayor = true;
  }
  ...
?>

then emit different html based on $isMayor:

<?php if ($isMayor) { ?>
  <div id="a">mayor</div>
<?php } else { ?>
  <div id="b">non-mayor</div>
<?php } ?>

...

<?php if ($isMayor) { ?>
  hey mayor
<?php } ?>
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