PHP not able to convert SQL result to string? [duplicate] - php

Question

This question already has answers here:
Object of class mysqli_result could not be converted to string
(5 answers)
Closed 1 year ago.
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('censored','censored','censored','db');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"db");
$sql="SELECT message FROM messages WHERE code = '".$q."'";
$result = mysqli_query($con,$sql);
$resultstring = (string)$result;
echo $resultstring;
mysqli_close($con);
?>
I am attempting to echo the result of the query to the user but when this PHP runs through ajax I get this error:
Recoverable fatal error: Object of class mysqli_result could not be
converted to string in D:\xampp\htdocs\getmessage.php on line 12
Now I don't understand this because I am already converting $result into a string.. Thanks!

Answer 1

mysqli_query() can not convert to string.
You must use mysqli_fetch for parse to array.
Example:
if ($result=mysqli_query($con,$sql))
{
// Fetch one and one row
while ($row=mysqli_fetch_row($result))
{
printf ("%s (%s)\n",$row[0],$row[1]);
}
}

Answer 2

First of all mysqli_query Performs a query on the database
it does not directly return a value
$query = mysqli_query($con,$sql);
so to do that use mysqli_fetch_assoc it will Fetch a result row as an associative array
$result = mysqli_fetch_assoc($query);
Then you can now get your value
$resultstring = $result['message'];
so your code should be like this
<?php
$q = intval($_GET['q']);
$con = mysqli_connect('censored','censored','censored','db');
if (!$con) {
die('Could not connect: ' . mysqli_error($con));
}
mysqli_select_db($con,"db");
$sql="SELECT message FROM messages WHERE code = '".$q."'";
$query = mysqli_query($con,$sql);
$result = mysqli_fetch_assoc($query);
$resultstring = $result['message'];
echo $resultstring;
mysqli_close($con);
?>

Answer 3

you need to convert the data(RS) into array !
$result = mysqli_fetch_assoc($result);
print_r($result);

Answer 4

As an result of your query on success You will get mysqli_result object. If You want to get received rows use fetch data function.
$result = mysqli_query($con,$sql);
$resultstring = $result->fetch_row()[0];
See docs: mysqli_query &
mysqli_result

Answer 5

I have resolved similar issue with following approach.
// Assign your sql statement
$sqlMasterSiteIds = "SELECT DISTINCT table_name FROM site WHERE gwid = 39 AND master_site_id NOT LIKE '0'";
// Run the query
$masterSiteIdsQuery = $this->db->query($sqlMasterSiteIds);
// Prepare container
$masterSiteIds = [];
// Fetch first row of query results
$first_row = $masterSiteIdsQuery->fetch_assoc();
// Assign first row to your prepared container
$masterSiteIds= $first_row['master_site_id'];
// Change the results into string comma separated so it can be used in next mysqli query or echoed
while($sub_row = $masterSiteIdsQuery->fetch_assoc()) {
$masterSiteIds = implode(',', array($masterSiteIds, $sub_row['master_site_id']));
}
// Now you can use $masterSiteIds which is a STRING of values comma separated
echo $masterSiteIds; // Output: '1024, 1142, 8492'
I hope it will help you or anybody else in need. I am not sure if it is the best approach but it worked well in my case as I had to change my results into STRING so I could use it in my next mysqli query with FIND_IN_SET command.
Happy coding!