php切换到mysqli：num_rows问题

发布时间：2020-12-13 17:44:09 所属栏目：PHP教程来源：网络整理

导读：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 =

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'];
    }

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.

解决方法

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,$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,$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() ;    

?>

（编辑：李大同）

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