Question

这是我第一次尝试使用SQLite，我试图从SQLite中检索数据但是没有成功，也没有抛出任何异常。

我的代码如下：

$sql_query = "SELECT * FROM `item` WHERE combo LIKE '" . $value . "' LIMIT 1;";
        try {
            /*** connect to SQLite database ***/
            $dbh = new PDO("sqlite:src/mydb.s3db");
            foreach ($dbh->query($sql_query) as $row) {
                $name = $row['name'];
                echo "name".$name; 
                $style = $row['style'];
                echo $style;
                $appr = $row['appr'];
                echo $appr;
                $style = $row['style'];
                echo $style;
            }

        } catch(PDOException $e) {
            //echo $e -> getMessage();
            echo "problem with DB";
            echo "<br>";
        }

请指导我完成这件事。谢谢。

//phpnifo();

PDO

PDO support enabled | PDO drivers mysql, sqlite, sqlite2

PDO_SQLITE

PDO Driver for SQLite 3.x enabled | PECL Module version (bundled) 1.0.1 $Id: pdo_sqlite.c 293036 2010-01-03 09:23:27Z sebastian $ SQLite Library 3.3.7

Answer 1

a）告诉PDO您希望它使用例外报告错误

$dbh = new PDO("sqlite::memory:");
$dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);

b）而不是将有效载荷的参与者放入sql语句中使用准备好的参数化语句

try {
    $stmt = $dbh->prepare( 'SELECT * FROM `item` WHERE combo LIKE ? LIMIT 1' );
    $stmt->execute( array($value) );
    foreach( $stmt as $row ) { ...

c）如果你仍然没有得到任何输出尝试无条件打印的代码。例如。 SELECT Count（*）将始终返回至少一条记录（如果没有错误发生）。

$stmt = $dbh->prepare( 'SELECT Count(*) as cnt FROM `item` WHERE combo LIKE ?' );
$stmt->execute( array($value) );
foreach( $stmt as $row ) {
    echo 'Count: ', $row['cnt'], "\n";
}
$stmt = null;

编辑：自包含示例

<?php
ini_set('display_errors', true);
error_reporting(E_ALL);

echo "start\n";

try {
    $dbh = new PDO("sqlite::memory:");
    $dbh->setAttribute(PDO::ATTR_ERRMODE, PDO::ERRMODE_EXCEPTION);
    $dbh->setAttribute(PDO::ATTR_EMULATE_PREPARES, false);
    setup($dbh);


    $stmt = $dbh->prepare('SELECT * FROM item WHERE combo LIKE ? LIMIT 1');
    $stmt->execute( array('comboB') );
    foreach( $stmt as $row ) {
        echo
            $row['name'], " ",
            $row['style'], " ",
            $row['appr'], "\n"
        ;
    }
}
catch(Exception $ex) {
    var_dump($ex);
}
echo "done.\n";

function setup($dbh) {
    $dbh->exec('
        CREATE TABLE item (
            combo TEXT,
            name TEXT,
            style TEXT,
            appr TEXT
        )
    ');
    $stmt = $dbh->prepare('INSERT INTO item (combo,name,style,appr) VALUES (?,?,?,?)');
    $stmt->execute( array('comboA','nameA','styleA','apprA') );
    $stmt->execute( array('comboB','nameB','styleB','apprB') );
    $stmt->execute( array('comboC','nameC','styleC','apprC') );
}

无法使用PHP从SQLite检索数据

1 个答案: