php切换到mysqli:num_rows问题

Hez*_*rac 4 php mysqli

我最近开始更新一些代码到MySQL改进的扩展,并且直到这一点已经成功:

// old code - works
$result = mysql_query($sql);
    if(mysql_num_rows($result) == 1){
    $row = mysql_fetch_array($result);
    echo $row['data'];
    }


// new code - doesn't work
$result = $mysqli->query($sql) or trigger_error($mysqli->error." [$sql]"); 
    if($result->num_rows == 1) {
    $row = $result->fetch_array();
    echo $row['data'];
    }
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如图所示,我试图使用面向对象的风格.我没有得到mysqli错误,vardump说没有数据......但db表中肯定有数据.

Mr.*_*cal 6

试试这个:

<?php

// procedural style

$host = "host";
$user = "user";
$password = "password";
$database = "db";

$link = mysqli_connect($host, $user, $password, $database);

IF(!$link){
    echo ('unable to connect to database');
}

ELSE {
$sql = "SELECT * FROM data_table LIMIT 1";
$result = mysqli_query($link,$sql);
    if(mysqli_num_rows($result) == 1){
    $row = mysqli_fetch_array($result, MYSQLI_BOTH);
    echo $row['data'];
    }
}
mysqli_close($link);


// OOP style 

$mysqli = new mysqli($host,$user, $password, $database);
$sql = "SELECT * FROM data_table LIMIT 1";
$result = $mysqli->query($sql) or trigger_error($mysqli->error." [$sql]"); /* I have added the suggestion from Your Common Sence */
    if($result->num_rows == 1) {
    $row = $result->fetch_array();
    echo $row['data'];
    }

    $mysqli->close() ;

// In the OOP style if you want more than one row. Or if you query contains more rows.    

$mysqli = new mysqli($host,$user, $password, $database);
$sql = "SELECT * FROM data_table";
$result = $mysqli->query($sql) or trigger_error($mysqli->error." [$sql]"); /* I have added the suggestion from Your Common Sence */
    while($row = $result->fetch_array()) {
     echo $row['data']."<br>";
    }

    $mysqli->close() ;    

?>
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  • 首先,为什么会有所作为?其次,你的`mysqli_query()`调用缺少它的第一个参数. (3认同)
  • 第三,`MYSQLI_BOTH`是默认参数,所以你什么都没改变.:) (2认同)