asked on
Warning: mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, string given in C:\....... on line 42
Hi guys,
I have this warning but I have not found what causes this error, please guys can you light me?
The error is in line 42 ---> while($row = mysqli_fetch_assoc($result
I have this warning but I have not found what causes this error, please guys can you light me?
The error is in line 42 ---> while($row = mysqli_fetch_assoc($result
$conn = mysqli_connect($DBhost, $DBuser, $DBpass, $DBname);
if (!$conn) {
die("No fue posible conectarse a la base de datos: " . mysqli_connect_error());
exit();
}else{
$sql = "SELECT * FROM documentos WHERE documentos.key = '$clavenum' or nombre = '$snombre' ORDER BY fecha DESC, hora DESC";
$result = mysqli_query($conn, $sql);
if (mysqli_num_rows($result) > 0) {
?> <table width="1000" border="0" align="center">
<tr>
<td> </td>
</tr>
<tr>
<td><span class="style2"></span></td>
</tr>
</table>
<table width="1000" border="0" align="center">
<tr>
<td width="150"><span class="style27 style2">Nombre de archivo:</span></td>
<td width="850"><span class="style37"><?php echo "$snombre"; ?></span></td>
</tr>
<tr>
<td> </td>
<td> </td>
</tr>
</table>
<table width="1000" border="0" align="center">
<tr bgcolor="#006A00">
<td width="9%"><div align="left" class="style35 style38">
<div align="center"><span class="style2">Fecha</span></div>
</div></td>
<td width="18%"><div align="left" class="style39">
<div align="center"><span class="style2">Usuario</span></div>
</div></td>
<td width="65%"><div align="left" class="style39">
<div align="center"><span class="style2">Descripción</span></div>
</div></td>
<td width="8%"><div align="left" class="style39">
<div align="center"><span class="style2">Anexo</span></div>
</div></td>
</tr> <?php
while($row = mysqli_fetch_assoc($result)) {
if ("-odd" == $odd) {
$odd = "";
echo "<tr bgcolor = #E6E6E6>";
}else{
$odd="-odd";
echo "<tr bgcolor = #ffffff>";
}
$cantidadcaracteres = strlen($row["rutaorigen"]);
if ($cantidadcaracteres > '0') {
$caracteres = ($cantidadcaracteres - 4);
$cualversion = substr($row["rutaorigen"], $caracteres, 1);
if ($cualversion == ".") {
$result = substr($row["rutaorigen"], -4);
}else{
$result = substr($row["rutaorigen"], -5);
}
echo ('<td><div class="style10"><font name="arial">' . $row["fecha"] . '</div></font></td>');
echo ('<td><div class="style10"><font name="arial">' . $row["usuario"] . '</div></font></td>');
echo ('<td><div class="style10"><font name="arial">' . $row["descripcion"] . '</div></font></td>');
echo ('<td><div class="style2" align="center"><font name="arial"><a href="\deptos\Uploads/' . $row["key"] . "$result" . '">' . "Anexo" . '</a></div></font></td>');
echo "</tr>";
}else{
echo ('<td><div class="style10"><font name="arial">' . $row["fecha"] . '</div></font></td>');
echo ('<td><div class="style10"><font name="arial">' . $row["usuario"] . '</div></font></td>');
echo ('<td><div class="style10"><font name="arial">' . $row["descripcion"] . '</div></font></td>');
echo ('<td><div class="style10"><font name="arial">' . "." . '</div></font></td>');
echo "</tr>";
}
}
?> </table> <?php
}else{
echo "No existen registros";
exit();
}
mysqli_close($conn);
}
?>
MySQL ServerPHP
Last Comment
Marco Gasi
Something is wrong in your query so the $result fariable is 'false' instead of a mysqli_result... Let me take a closer look
Try to get the error:
$sql = "SELECT * FROM documentos WHERE documentos.key = '$clavenum' or nombre = '$snombre' ORDER BY fecha DESC, hora DESC";
$result = mysqli_query($conn, $sql);
if (!$result)
{
echo("Error description: " . mysqli_error($con));
}
else
{
...
ASKER
I tried but displays the same error only on line 42
$sql = "SELECT * FROM documentos WHERE documentos.key = '$clavenum' or nombre = '$snombre' ORDER BY fecha DESC, hora DESC";
$result = mysqli_query($conn, $sql);
if (!$result)
{
echo("Error description: " . mysqli_error($con));
$sql = "SELECT * FROM documentos WHERE documentos.key = '$clavenum' or nombre = '$snombre' ORDER BY fecha DESC, hora DESC";
$result = mysqli_query($conn, $sql);
if (!$result)
{
echo("Error description: " . mysqli_error($con));
Try to echo the sql query to see it is what you expect it to be:
$sql = "SELECT * FROM documentos WHERE documentos.key = '$clavenum' or nombre = '$snombre' ORDER BY fecha DESC, hora DESC";
echo $ql;
exit;
$sql = "SELECT * FROM documentos WHERE documentos.key = '$clavenum' or nombre = '$snombre' ORDER BY fecha DESC, hora DESC";
echo $ql;
exit;
ASKER
The echo displays this:
SELECT * FROM documentos WHERE documentos.key = '05115' or nombre = 'Solicitud de Apoyo logistico' ORDER BY fecha DESC, hora DESC
and yes, it is what i'm looking for
SELECT * FROM documentos WHERE documentos.key = '05115' or nombre = 'Solicitud de Apoyo logistico' ORDER BY fecha DESC, hora DESC
and yes, it is what i'm looking for
Ok, this is a stupid question, I know, but... are you sure the record i present in the database?
ASKER
Yes it exist, actually it displays one record
ASKER
And there are 5 records that must be displayed but displays only one and the warning line
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ASKER
wow, wow, wow
Oh my Goood Marco, you got it.
I'll take note for the next one :)
Thank you so much, you made may day.
Oh my Goood Marco, you got it.
I'll take note for the next one :)
Thank you so much, you made may day.
Lol, glad to help you. On to hte next.