PHP全文搜索无法正常工作-Java 学习之路

我正在尝试对列执行FULLTEXT搜索，如果第一个返回零行，则进一步在布尔模式下搜索FULLTEXT . 如果我在命令行手动输入SQL查询到MySQL，我得到了我期望的结果，但这段代码返回一个空数组到我的JS . 我要去的任何想法？

$con = mysqli_connect( $db_url, $db_user, $db_pwd, $db );
if( !$con ) {
error_log("Connection failed!");
die();
}

$sql = "SELECT problem, note_text, doctor, entry_date FROM notes WHERE mrn=? AND MATCH(note_text) AGAINST(?)";

$stmt = mysqli_prepare( $con, $sql );
mysqli_stmt_bind_param( $stmt, 'is', $_POST['mrn'], $_POST['search_text'] );
$success = mysqli_stmt_execute( $stmt );

if( mysqli_stmt_num_rows($stmt) == 0 ) { // Then try fulltext search in boolean mode

    $sql = "SELECT problem, note_text, doctor, entry_date FROM notes WHERE mrn=? AND MATCH(note_text) AGAINST(? IN BOOLEAN MODE)";

    $stmt = mysqli_prepare( $con, $sql );
    mysqli_stmt_bind_param( $stmt, 'is', $_POST['mrn'], $_POST['search_text'] );
    $success = mysqli_stmt_execute( $stmt );
} 

    mysqli_bind_result( $stmt, $problem, $note_text, $doctor, $date );
    $notes = array();

    for( $i = 0; mysqli_stmt_fetch($stmt) == TRUE; ++$i ) {
        $notes[$i] = array( 'problem' => $problem, 'note' => $note_text, 'doctor' => $doctor, 'date' => $date );
    }

    $json = json_encode($notes);
    echo $json;

1 回答

0
好的，我想我找到了答案 . 在我使用第二个调用之前，我需要关闭第一个STATEMENT $ stmt
```
mysqli_stmt_close($stmt)
```
刚过
```
if( mysqli_stmt_num_rows($stmt) == 0 )
```
干脆做
```
$stmt = mysqli_prepare( $con, $sql );
```
不会给你一个新的声明，并覆盖以前的变量！
回复于 2024-05-17T18:21:11+08:00

PHP全文搜索无法正常工作

1 回答

相关问题