Question

我得到了这样的警告＆＃34; mysql_fetch_assoc（）期望参数1是资源＆＃34;，我该如何解决这个问题？谢谢！

global $wpdb;

$con=mysqli_connect("localhost","root","");
mysqli_select_db('search');

$search = mysql_real_escape_string(trim($_POST['searchquery']));
$sqlCommand = $wpdb->get_results("SELECT * FROM `wp_doctors` WHERE `name` LIKE '%searchquery%'");


while($row = mysql_fetch_assoc($sqlCommand)){
    $id = $row['id'];
    $name = $row['name'];
    $spec = $row['spec'];
    echo "Nume: $name<br />Specializare: $spec";
}

Answer 1

没有必要这样做，

$row = mysql_fetch_assoc($sqlCommand)

$ wpdb-＆gt; get_results已经完成了var_dump $sqlCommand的工作，你会看到错误。

Answer 2

可以使用get_results从数据库中提取多行结果。该函数将整个查询结果作为数组返回，

因此，您可以在代码中进行此更改：

global $wpdb;

$con=mysqli_connect("localhost","root","");
mysqli_select_db('search');

$search = mysql_real_escape_string(trim($_POST['searchquery']));
$sqlCommand = $wpdb->get_results("SELECT * FROM `wp_doctors` WHERE `name` LIKE '%searchquery%'");

foreach ( $sqlCommand as $result) 
{
    echo $result->id;
        echo $result->name;
        echo $result->spec;
}

自定义搜索php为wordpress

2 个答案: