php – mysql.proc的列计数错误.预计20,发现16.表可能已损坏
发布时间:2020-12-13 16:34:52 所属栏目:PHP教程 来源:网络整理
导读:我正在使用000webhost.com,我正在使用phpMyAdmin.当我运行我的 PHP脚本时,我从mysql得到这个错误,标题如下: Column count of mysql.proc is wrong. Expected 20,found 16. The table is probably corrupted. 有没有解决方案?谢谢. 编辑:这是我的代码 ?php
我正在使用000webhost.com,我正在使用phpMyAdmin.当我运行我的
PHP脚本时,我从mysql得到这个错误,标题如下:
有没有解决方案?谢谢. 编辑:这是我的代码 <?php $username="usrname"; $password="passwd"; $database="a1xxxxx_mydb"; $host="mysqlxx.000webhost.com"; mysql_connect($host,$username,$password); @mysql_select_db($database) or die( "Unable to select database"); if (isset($_GET["userLatitude"]) && isset($_GET["userLongitude"])) { $userLatitude=$_GET['userLatitude']; $userLongitude=$_GET['userLongitude']; $result = mysql_query("SELECT locationName,( 6371 * acos( cos( radians(floatval( $userLatitude) )) * cos( radians( locationLatitude ) ) * cos( radians( locationLongitude ) - radians( floatval($userLatitude)) ) + sin( radians(floatval($userLongitude)) ) * sin( radians( locationLatitude) ) ) ) AS distance FROM Location HAVING distance < 2 ORDER BY distance LIMIT 0,20") or die(mysql_error()); echo $result; // check for empty result if (mysql_num_rows($result) > 0) { // looping through all results // products node $response["Location"] = array(); while ($row = mysql_fetch_array($result)) { // temp user array $product = array(); $product["locationName"] = $row["locationName"]; $product["locationInfo"] = $row["locationInfo"]; $product["locationLatitude"] = $row["locationLatitude"]; $product["locationLongitude"] = $row["locationLongitude"]; $product["locationPic"] = $row["locationPic"]; $product["city"] = $row["city"]; // push single product into final response array array_push($response["Location"],$product); } // success $response["success"] = 1; // echoing JSON response echo json_encode($response); } else { // no products found $response["success"] = 0; $response["message"] = "No products found"; // echo no users JSON echo json_encode($response); } } else { // required field is missing $response["success"] = 0; $response["message"] = "Required field(s) is missing"; // echoing JSON response echo json_encode($response); } mysql_close(); ?>
我也有这个错误.我通过运行来修复它
mysql_upgrade -u root -p 另外,通过运行重新启动mysql服务 service mysqld restart (编辑:李大同) 【声明】本站内容均来自网络,其相关言论仅代表作者个人观点,不代表本站立场。若无意侵犯到您的权利,请及时与联系站长删除相关内容! |