PHP mySQLi query returning data but not displaying it
Question
Good day, I have done extensive research on this issue but unfortunately none of the related issues solved my problem.
Here I have a very basic PHP mySQLi db connection. The connection succeeds and so does the query that is run on the table. The issue is that the the result set will not display. All of my references are correct and when I check to see if the result set is populated, it is. I believe the issue is with my while block but no errors are returned when this is run. Thank you for your time
<?php
$db = mysqli_connect('localhost','root','','securitour') //connection to the database
or die('Error connecting to MySQL server.');
?>
<html>
<head>
</head>
<body>
<?php
$query = "SELECT * FROM location"; //The SQL query
mysqli_query($db, $query) or die('Error querying database.');
$result = mysqli_query($db, $query); //query the table an store the result set in a variable
$row = mysqli_fetch_array($result); //create an array and store the records of the result set in it
if (mysqli_num_rows($result) != 0) //to check if the result set contains data
{
echo "results found"; //THIS is what is returned.
}
else
{
echo "results not found";
}
while ($row = $result->fetch_assoc()) //itterate through the array and display the name column of each record
{
echo $row['name'];
}
mysqli_close($db);
?>
</body>
</html>
2 answers
solution1
1 2018-11-02 10:49:08
You don't need to run mysqli_query() twice. and you need to use mysqli_fetch_assoc for associative array
<?php
$db = mysqli_connect('localhost','root','','securitour') or die('Error connecting to MySQL server.');
?>
<html>
<head>
</head>
<body>
<?php
$query = "SELECT * FROM location"; //The SQL query
$result = mysqli_query($db, $query) or die('Error querying database.'); //query the table an store the result set in a variable
$row = mysqli_fetch_assoc($result); //create an array and store the records of the result set in it
if (mysqli_num_rows($result) != 0) //to check if the result set contains data
{
echo "results found"; //THIS is what is returned.
} else {
echo "results not found";
}
foreach ( $row as $name=>$val) {
echo $name . ':' . $val . '<br>';
}
mysqli_close($db);
?>
</body>
</html>
solution2
1 ACCPTED 2018-11-02 10:53:31
Lots of things not right here .
You are processing the mysqli_query() function twice - there is no need.
You are selecting all fields in your SQL query (SELECT *). You should select fields by name.
- You're interchanging between procedure and class-based MySQLi - You should stick to one or the other.
Try this instead:
<?php
$db = mysqli_connect('localhost','root','','securitour') //connection to the database
or die('Error connecting to MySQL server.');
?>
<html>
<head>
</head>
<body>
<?php
$query = "SELECT name FROM location"; //The SQL query
$result = mysqli_query($db, $query) or die('Error querying database'); //query the table an store the result set in a variable
if(mysqli_num_rows($result) > 0){
echo "Results found!";
while($row = mysqli_fetch_array($result)){ //create an array and store the records of the result set in it
echo $row['name'];
}
} else {
echo "results not found";
}
mysqli_close($db);
?>
</body>
</html>
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