[SOLVED] Can't get PHP to print a result of an sql query.

[heldenbrau]

What is wrong with this, I just want it to print what's in the rating field, but nothing happens.

<?php

   $mysqli = new mysqli("localhost", "user", "password", "user");
    if ($mysqli === false) {
      die("ERROR: Could not connect to database. " . mysqli_connect_error());
    }

$sql = "SELECT rating from users where username = \"Chris\"";

if ($result = $mysqli->query($sql)){
echo "$result";
} else {
echo "no results";
$result->close();
$mysqli->close();
?>

[lonewolf217]

SQL requires single quotes for text fields (i.e. around 'Chris'), double quotes wont work from my understanding.

 

your bigger problem though is you do not close the bracket for your else statement which is probably causing the page to not show.

 

You need to turn on error reporting to see the errors for stuff like this

[heldenbrau]

Thanks that has got the page showing, but now it says no results.  I have changed the script to

 

<?php

   $mysqli = new mysqli("localhost", "user", "password", "user");
    if ($mysqli === false) {
      die("ERROR: Could not connect to database. " . mysqli_connect_error());
    }

$sql = "SELECT rating from users where username = \'Chris\'";

if ($result = $mysqli->query($sql)){
echo "$result";
} else {
echo "no results";
}

$result->close();
$mysqli->close();
?>

[lonewolf217]

take out the slashes

<?php
$sql = "SELECT rating from users where username = 'Chris'";

 

I am not entirely sure if this will work with your class structure, but try it

 

<?php
$result = $mysqli->query($sql) or die(mysql_error());

if it doesn't, then you should just take exactly what you have for your $sql statement and throw it into your sql query engine and try to run it manually to see what the errors are

[J.Daniels]

If you are using single quotes inside of double quotes, you don't have to escape them.

 

$sql = "SELECT rating from users where username = 'Chris'";

 

As a side note, MySQL works fine with double quotes.

[lonewolf217]

maybe its just MSSQL, because when I run this

 

select * from table where action like "%login%"

 

i get

"Invalid column name '%login%'"

[heldenbrau]

It works fine when I type it directly into mysql.  I have taken out the slashes and it goes back to nothing happening. 

 

<?php

   $mysqli = new mysqli("localhost", "user", "password", "user");
    if ($mysqli === false) {
      die("ERROR: Could not connect to database. " . mysqli_connect_error());
    }

$sql = "SELECT rating from users where username = 'Chris'";

if ($result = $mysqli->query($sql)){
echo "$result";
} else {
echo "no results";
}

$result->close();
$mysqli->close();
?>

 

 

[Philip]

<?php

   $mysqli = new mysqli("localhost", "user", "password", "user");
    if ($mysqli === false) {
      die("ERROR: Could not connect to database. " . mysqli_connect_error());
    }

$sql = "SELECT rating from users where username = 'Chris'";

if ($result = $mysqli->query($sql)){
echo "$result";
} else {
echo "no results, but here is the error: ",$mysqli->error;
}

$result->close();
$mysqli->close();
?>

[lonewolf217]

also, put this at the top of your page to view any other errors

 

<?php
error_reporting(E_ALL);
ini_set("display_errors", 1); 

[heldenbrau]

I put that at the top and I get this error

 

Catchable fatal error: Object of class mysqli_result could not be converted to string

[J.Daniels]

That is because you are trying to echo out the results of the query:

 

$result = $mysql->query($sql);
if ($row = $result->fetch_assoc()){
    echo $row['rating'];
} else {
    echo "no results, but here is the error: ",$mysqli->error;
}

[heldenbrau]

After replacing the code with the above it now says undefined variable and Fatal error: Call to a member function query() on a non-object.

[lonewolf217]

change $mysql->query() to $mysqli->query()

[heldenbrau]

Yay, that's done it  , wish I understood it though.  It'll come to me.

 

Thanks for that.

[heldenbrau]

Now I need to do the same thing, but with a date field.  The fetch_assoc()) doesn't work with the date field.

[lonewolf217]

post the code and the error you are getting.

 

did  you change your original SQL query, because there you were only retrieving the one field, "rating" so if you try to grab the date field from that it would throw an error

[heldenbrau]

don't worry, my fault, was using the wrong field name.  It does work with dates.