Question

我似乎无法解决此问题，尽管我认为这与我的语法有关。任何帮助将不胜感激。

“解析错误：语法错误，意外的'，'，期望第12行中的']'是显示的错误。

?php
if ( ! empty( $_POST ) ) {

    $mysqli = new mysqli('localhost', '111', '111', '111' );

    if ( $mysqli->connect_error ){
        die('connect error: ' . $mysqli->connect_errno. ': ' . $mysqli->connect_error );
    }

    $sql = "INSERT INTO processcv ( Name, Surname, Tel, Email, Message, CVtype) VALUES 
    ( '{$mysqli->cubrid_real_escape_string($_POST['Name'])}','
    ( '{$mysqli->cubrid_real_escape_string($_POST['Surname])}','
    ( '{$mysqli->cubrid_real_escape_string($_POST['Tel'])}','
    ( '{$mysqli->cubrid_real_escape_string($_POST['Email'])}','
    ( '{$mysqli->cubrid_real_escape_string($_POST['Message'])}',
    ( '{$mysqli->cubrid_real_escape_string($_POST['CVtype'])}' )";
    $insert = $mysqli->query($sql);

    if ( $insert ){
    echo "success! Row ID: {$mysqli->insert_id}";
    } else{
    die("error: {$mysqli->errno} : {$mysqli->error}");
    }

    $mysqli->close();

}
?>

Answer 1

他们的语法错误意外地使用了'和称为变量的方式，请在下面检查

<?php
if ( ! empty( $_POST ) ) {

    $mysqli = new mysqli('localhost', '111', '111', '111' );

    if ( $mysqli->connect_error ){
        die('connect error: ' . $mysqli->connect_errno. ': ' . $mysqli->connect_error );
    }

    $sql = "INSERT INTO processcv ( Name, Surname, Tel, Email, Message, CVtype) VALUES 
    ( '".$mysqli->cubrid_real_escape_string($_POST['Name'])."'),
    ( '".$mysqli->cubrid_real_escape_string($_POST['Surname'])."'),
    ( '".$mysqli->cubrid_real_escape_string($_POST['Tel'])."'),
    ( '".$mysqli->cubrid_real_escape_string($_POST['Email'])."'),
    ( '".$mysqli->cubrid_real_escape_string($_POST['Message'])."'),
    ( '".$mysqli->cubrid_real_escape_string($_POST['CVtype'])."')";
    $insert = $mysqli->query($sql);

    if ( $insert ){
    echo "success! Row ID: {$mysqli->insert_id}";
    } else{
    die("error: {$mysqli->errno} : {$mysqli->error}");
    }

    $mysqli->close();

}
?>

使用php将用户输入的数据存储到数据库中时出错

1 个答案: