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Problema PHP


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  • 1 month later...

1. Nu folosi

SELECT *

unde nu ai nevoie. Chestia asta nu e ok, e recomandat sa selectezi doar ce ai nevoie, in cazul tau:

SELECT imagine FROM `oferte` WHERE `id` = ?

2. Ai 2 posibilitati ca sa faci echo la o variabila in content html.

  • echo "<img src = {$url}>";
  • echo "<img src = ".$url.">";

 

 

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