mysqli_fetch_assoc() expects parameter 1 to be mys

2020-01-24 13:12发布

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问题:

This question already has an answer here:

mysqli_fetch_assoc() expects parameter / Call to a member function bind_param() errors. How to get the actual mysql error and fix it? (1 answer)

Closed 2 years ago.

I am pretty new to PHP and MySQL and I just can't figure this one out. I have searched all around the forum but haven't found an answer I can make sense of. I originally was using mysql_fetch_assoc() but I could only search numbers and I received errors when searching for letters as well. I hope I am on the right track here. Thank you in advance for all your help!

$con = mysqli_connect($hostname,$username,$password) or die ("<script language='javascript'>alert('Unable to connect to database')</script>");
mysqli_select_db($con, $dbname);

if (isset($_GET['part'])){
    $partid = $_GET['part'];
    $sql = 'SELECT * 
        FROM $usertable 
        WHERE PartNumber = $partid';

    $result = mysqli_query($con, $sql);
    $row = mysqli_fetch_assoc($result);

    $partnumber = $partid;
    $nsn = $row["NSN"];
    $description = $row["Description"];
    $quantity = $row["Quantity"];
    $condition = $row["Conditio"];
}

回答1:

This happens when your result is not a result (but a "false" instead). You should change this line

$sql = 'SELECT * FROM $usertable WHERE PartNumber = $partid';

to this:

$sql = "SELECT * FROM $usertable WHERE PartNumber = $partid";

because the " can interprete $variables while ' cannot.

Works fine with integers (numbers), for strings you need to put the $variable in single quotes, like

$sql = "SELECT * FROM $usertable WHERE PartNumber = '$partid' ";

If you want / have to work with single quotes, then php CAN NOT interprete the variables, you will have to do it like this:

 $sql = 'SELECT * FROM '.$usertable.' WHERE string_column = "'.$string.'" AND integer_column = '.$number.';

回答2:

mysqli_fetch_assoc() expects parameter 1 to be mysqli_result, boolean given

This means that the first parameter you passed is a boolean (true or false).

The first parameter is $result, and it is false because there is a syntax error in the query.

" ... WHERE PartNumber = $partid';"

You should never directly include a request variable in a SQL query, else the users are able to inject SQL in your queries. (See SQL injection.)

You should escape the variable:

" ... WHERE PartNumber = '" . mysqli_escape_string($conn,$partid) . "';"

Or better, use Prepared Statements.

回答3:

You are single quoting your SQL statement which is making the variables text instead of variables.

$sql = "SELECT * 
    FROM $usertable 
    WHERE PartNumber = $partid";

回答4:

Mysqli makes use of object oriented programming. Try using this approach instead:

function dbCon() {
        if($mysqli = new mysqli('$hostname','$username','$password','$databasename')) return $mysqli; else return false;
}

if(!dbCon())
exit("<script language='javascript'>alert('Unable to connect to database')</script>");
else $con=dbCon();

if (isset($_GET['part'])){
    $partid = $_GET['part'];
    $sql = "SELECT * 
        FROM $usertable 
        WHERE PartNumber = $partid";

    $result=$con->query($sql_query);
    $row = $result->fetch_assoc();

    $partnumber = $partid;
    $nsn = $row["NSN"];
    $description = $row["Description"];
    $quantity = $row["Quantity"];
    $condition = $row["Conditio"];
}

Let me know if you have any questions, I could not test this code so you might need to tripple check it!