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[php] Versteckter SELECT-Fehler

[php] Versteckter SELECT-Fehler

Stefan Landgraf stefan_(at)_living-source.com
Tue, 6 Jun 2000 14:23:51 +0200 

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Hi Domink,
  $query =3D "SELECT * FROM eintrag
      WHERE counter LIKE '$added'";
   $mysql_result =3D mysql_query($query, $mysql_link);

Das sollte funzen.

Gru=DF, Stefan

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<!DOCTYPE HTML PUBLIC "-//W3C//DTD HTML 4.0 Transitional//EN">
<HTML><HEAD>
<META content=3D"text/html; charset=3Diso-8859-1" =
http-equiv=3DContent-Type>
<META content=3D"MSHTML 5.00.2919.6307" name=3DGENERATOR>
<STYLE></STYLE>
</HEAD>
<BODY bgColor=3D#ffffff>
<DIV><FONT face=3DArial size=3D2>Hi Domink,</FONT></DIV>
<BLOCKQUOTE=20
style=3D"BORDER-LEFT: #000000 2px solid; MARGIN-LEFT: 5px; MARGIN-RIGHT: =
0px; PADDING-LEFT: 5px; PADDING-RIGHT: 0px">
  <DIV>$query =3D "SELECT * FROM =
eintrag<BR>&nbsp;&nbsp;&nbsp;&nbsp;WHERE counter=20
  LIKE '$added'";<BR>&nbsp;$mysql_result =3D mysql_query($query,=20
  $mysql_link);</DIV>
  <DIV>&nbsp;</DIV></BLOCKQUOTE>
<DIV><FONT face=3DArial size=3D2>Das sollte funzen.</FONT></DIV>
<DIV>&nbsp;</DIV>
<DIV><FONT face=3DArial size=3D2>Gru=DF, =
Stefan</FONT></DIV></BODY></HTML>

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