javascript - 第 6 行 mysql_fetch_array 错误

Question

mysql_fetch_array() expects parameter 1 to be resource, boolean given in

<?php
    mysql_connect ("localhost", "cab","a321")  or die (mysql_error());
    mysql_select_db ("ppwxpjey_mcidb");
    $termOrd = $_POST['termOrd'];
    $sql = mysql_query("select * from booking where order_no like '%$termOrd%'");
    while ($row = mysql_fetch_array($sql)){
        echo "<table width='1000' border='2' align='center' style='background-color:#FFFFFF;border-collapse:collapse;border:2px solid #6699FF;color:#000000'><tr><th>ORDER NO</th><th>NAME</th><th>MOBILE</th><th>FROM PLACE</th><th>TO PLACE</th><th>JOURNEY DATE</th><th>JOURNEY TIME</th><th>PERSON</th><th>BOOKING TIME</th></tr>";
        echo "<tr><td>".$row["ORDER_NO"]."</td><td>".$row["NAME"]."</td><td>".$row["MOBILE"]."</td><td>".$row["FROM_PLACE"]."</td><td>".$row["TO_PLACE"]."</td><td>".$row["JOURNEY_DATE"]."</td><td>".$row["JOURNEY_TIME"]."</td><td>".$row["PERSON"]."</td><td>".$row["UPDATE_TIME"]."</td></tr>";
        echo '<br/>';
        }
    ?>

Answer 1

您必须在将查询传入 mysql_fetch_array 之前运行该查询它期望资源作为参数。所以像这样改变你的代码，

$sql = mysql_query("select * from booking where order_no like '%$termOrd%'");

$result = mysql_query($query) or die(mysql_error()); 
// This is where you're getting resource or throwing SQL error.

while ($row = mysql_fetch_array($result)){
        // YOUR LOGIC.
}

Warning mysql_query, mysql_fetch_array,mysql_connect etc.. extensions were deprecated in PHP 5.5.0, and it was removed in PHP 7.0.0.
Instead, the MySQLi or PDO_MySQL extension should be used.
Source : http://php.net/manual/en/function.mysql-query.php

因此请尽快更新您的代码。