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Elseif statement only displaying the result of the 'if' part

carter90

By carter90
in PHP Coding Help

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carter90 Newbie

carter90

Posted
Posted

I have this code below which is linked to my mysql database. For some reason it is only showing the price for the if part of the statement?

<html>
<body>
<?php

$mysqli = new mysqli("localhost", "root", "", "insuredcars");
if ($_POST['formcar'] == '1' && $_POST['formage'] == '18' &&  $_POST['formNCD'] == '0' && $_POST['formPoints'] == '0' )
{
    $query = $mysqli->query("SELECT * FROM insurance WHERE insuranceid = '1'");
    while($row = mysqli_fetch_assoc($query))
    {
        echo "<tr>";
        echo "<td> The price for insurance will be " . $row['insuranceprice'] . "</td>";
        echo "</tr>";
		
	}
     echo "</table>";   
}
elseif ($_POST['formcar'] == '2' && $_POST['formage'] == '18' &&  $_POST['formNCD'] == '0' && $_POST['formPoints'] == '0' )
{
    $query = $mysqli->query("SELECT * FROM insurance WHERE insuranceid = '2'");
	
    echo "<tr>";
    echo "<td> The price for insurance will be " . $row['insuranceprice'] . "</td>";
    echo "</tr>";
	

echo "</table>";
}
elseif ($_POST['formcar'] == '3' && $_POST['formage'] == '18' &&  $_POST['formNCD'] == '0' && $_POST['formPoints'] == '0' )
{
    $query = $mysqli->query("SELECT * FROM insurance WHERE insuranceid = '5'");
	
    echo "<tr>";
    echo "<td> The price for insurance will be " . $row['insuranceprice'] . "</td>";
    echo "</tr>";
	
	echo "</table>";
}

elseif ($_POST['formcar'] == '4' && $_POST['formage'] == '18' &&  $_POST['formNCD'] == '0' && $_POST['formPoints'] == '0' )
{
    $query = $mysqli->query("SELECT * FROM insurance WHERE insuranceid = '6'");
	
    echo "<tr>";
    echo "<td> The price for insurance will be " . $row['insuranceprice'] . "</td>";
    echo "</tr>";
	
	echo "</table>";
}

?>
</body>
</html>
DavidAM Member

DavidAM

Posted
Posted

You never retrieve the results of the query in any of the ELSEIF blocks. mysqli_query only executes the query, you need to use mysqli_fetch_assoc after that to get the data.

 

And, by the way, if are only selecting one row and only processing one row, you do not need the while loop. You can just do the fetch.

carter90 Newbie

carter90

Posted
  • Author
Posted

Thanks for that I have added the code in below the elseif statements:

 

I wasn't sure what put instead of while as it comes up with a syntax error if I just take it out.

 

while ($row = mysql_fetch_assoc($query))

 

 

Unfortunately now it is coming up with an error:

 

Warning: mysql_fetch_assoc() expects parameter 1 to be resource, object given in search.php on line 9

 

Sorry if I am being an idiot.

DavidAM Member

DavidAM

Posted
Posted

You are using mysqli in your original code. You can NOT mix mysql calls with that. You must use mysqli_fetch_assoc().

 

If you are expecting and wanting only ONE row. You just need the $row = mysqli_fetch_assoc($query); without the WHILE () { and, of course, remove the closing brace afterwards.

 

$query = $mysqli->query("SELECT * FROM insurance WHERE insuranceid = '1'");
$row = mysqli_fetch_assoc($query); #MAD# Added
#MAD# while($row = mysqli_fetch_assoc($query)) # Removed
#MAD# { # Removed
echo "<tr>";
echo "<td> The price for insurance will be " . $row['insuranceprice'] . "</td>";
echo "</tr>";
#MAD# } # Removed
echo "</table>";

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