PHP变量未显示值
我将MySQL查询值存储在PHP变量中,但它不显示数据。备注:数据在MySQL表列中可用PHP变量未显示值,php,mysql,sql,select,Php,Mysql,Sql,Select,我将MySQL查询值存储在PHP变量中,但它不显示数据。备注:数据在MySQL表列中可用 <?php $cmsca= mysql_query("SELECT SUM(qa_effort) FROM tbl_uat WHERE product='CAP'"); while ($cresulta = mysql_fetch_array ($cmsca)) $arra[0] = $cresulta[0]; echo $arra[0]; ?>
<?php
$cmsca= mysql_query("SELECT SUM(qa_effort) FROM tbl_uat WHERE product='CAP'");
while ($cresulta = mysql_fetch_array ($cmsca))
$arra[0] = $cresulta[0];
echo $arra[0];
?>
试试这个
<?php
$cmsca= mysql_query("SELECT SUM(qa_effort) as sums FROM tbl_uat WHERE product='CAP'");
while ($cresulta = mysql_fetch_array($cmsca))
{
echo $cresulta['sums'];
}
?>
首先,不要使用mysql\u query,-它已被弃用,请改用。
接下来,在运行查询之前,您需要连接到数据库
$mysqli = new mysqli("localhost", "my_user", "my_password", "world");
/* check connection */
if ($mysqli->connect_errno) {
printf("Connect failed: %s\n", $mysqli->connect_error);
exit();
}
$result = mysqli->query("SELECT SUM(qa_effort) as sums FROM tbl_uat WHERE product='CAP'");
while ($row = $result->fetch_array()) {
var_dump($row);
}
$mysqli->close();
?>
试试这个怎么样:
<?php
$arra = array();
$cmsca= mysql_query("SELECT SUM(qa_effort) FROM tbl_uat WHERE product='CAP'");
while ($row = mysql_fetch_array ($cmsca))
$arra = $row;
print_r($arra);
?>
您是否尝试过使用var_dump($cresulta[0])代码>查看里面有什么?有时echo
无法显示某些值(如false
),请确保您的查询返回结果,否则您的操作是正确的。do:var_dump($cresulta)inside while loopAnswer可以,但解释会使答案更好+1。@Farhad事实是我不知道该解释什么:),这是基本的,他混合了mysql\u fetch\u array
和mysql\u fetch\u num
。