2015-12-27 22:14 阅读 38


Im not really sure how to use mysqli inside of classes as I'm used to procedural programming but I understand where OOP is beneficial. So I'm trying to create an update function in my class using mysqli but it does not seem to be working. I also cannot figure out how to get it to echo out errors. It just tells me this:

mysqli_error() expects parameter 1 to be mysqli

This is the object:

$mysqli = new mysqli('localhost', 'root', '', 'ams');
$part = new Part($mysqli);

Here is my class:

class Part {

    protected $mysqli;

    // Editable Part table titles
    protected $editable = array('Desc', 'SDQty', 'IAQty', 'IsActive');

    function __construct($mysqli) {
        $this->mysqli = $mysqli;

    function update($data = array(), $partID) {

        $r_errors = 0;

        // Data Arrays
        $updateArray = array();

        // Loop Though all the column titles
        foreach ( $data as $column => $value ) {

            // If the column value is set
            if ( ( isset($column) && !empty($column) ) && ( isset($value) && !empty($value) ) ) {

                // If the item to be updated is allowable from the allowable array || add it to the column array
                if ( in_array($column, $this->editable) ) {
                    $updateArray[] = "$column = $value";


        } // End of foreach loop

        // Convert Data Arrays to strings
            $updateData = implode(', ', $updateArray);

        // Create the SQL string
            $_query = "UPDATE parts WHERE PartID = $partID SET $updateData";
            $result = $this->mysqli->query($_query);

            if ( !$result ) {
                echo mysqli_error($result);

It will not run the query and Im kind of a beginner at this OOP with php and mysql so any help is greatly appreciated.

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2条回答 默认 最新

  • 已采纳
    douxiong3245 douxiong3245 2015-12-27 22:23

    You have your query wrong, instead of:

    UPDATE parts WHERE PartID = $partID SET $updateData;

    It should be (note the order of things):

    UPDATE parts SET column_name = 'column_data' WHERE PartID = $partID;

    Take a look at UPDATE Syntax for more information.

    About the error :

    mysqli_error() expects parameter 1 to be mysqli

    Since $mysqli is the variable that holds your mysqli, that's the one that should be used:

    echo mysqli_error($mysqli);

    NOTE: I don't know where this data is coming from, but in case it's user input, you should definitely take a look at How can I prevent SQL-injection in PHP?

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  • dongqijuan3786 dongqijuan3786 2015-12-27 22:19

    The line echo mysqli_error($result); is incorrect. You are attempting to pass the result set as the parameter.

    What you should have is;

    echo $mysqli->error();

    to keep the correct OO syntax.

    You could have mysqli_error($mysqli), but that is procedural syntax.

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