<?php
$sqls = mysql_query("SELECT weight FROM $usertablestats")
or die ("Query failed: " . mysql_error() . " Actual query: " . $query);
$ct = mysql_query ("COUNT * '$sqls'");
if ($ct > 0) {
while ($row = mysql_fetch_array($sqls));{
$weight = $row["weight"];
echo "C" . $weight;
}}
else {
echo "No stats found";
}
?>
即使我在表格中有数据,也会输出“找不到统计数据”。
<?php
$sqls = mysql_query("SELECT weight FROM $usertablestats")
or die ("Query failed: " . mysql_error() . " Actual query: " . $query);
$ct = mysql_num_rows($sqls);
if ($ct > 0) {
while ($row = mysql_fetch_array($sqls));{
$weight = $row["weight"];
echo "C" . $weight;
}}
else {
echo "No stats found";
}
?>
这不会返回任何内容。根本没有回音。
我已经通过以下方式检查是否正在访问:
<?php
$sqls = mysql_query("SELECT weight FROM $usertablestats")
or die ("Query failed: " . mysql_error() . " Actual query: " . $query);
$row = mysql_fetch_array($sqls);
echo $row;
?>
它确实返回了第一个条目。
答案 0 :(得分:1)
你有分号:
while ($row = mysql_fetch_array($sqls));{ //should be while ($row = mysql_fetch_array($sqls)){
是否导致问题
答案 1 :(得分:0)
试试这个:
<?php
$sqls = mysql_query("SELECT weight FROM $usertablestats")
or die ("Query failed: " . mysql_error() . " Actual query: " . $query);
int $ct = 0;
while ($row = mysql_fetch_array($sqls)){
$weight = $row["weight"];
echo "C" . $weight;
$ct++;
}
if ($ct == 0) {
echo "No stats found";
}
?>
如果不起作用,请确保$usertablestats
具有正确的值。
答案 2 :(得分:0)
试试这个。
while ($row = mysql_fetch_array($sqls,MYSQL_ASSOC)){
$weight = $row["weight"];
echo "C" . $weight;
$ct++;
}
答案 3 :(得分:0)
<?php
$sqls = mysql_query("SELECT * FROM $usertablestats")
or die ("Query failed: " . mysql_error() . " Actual query: " . $query);
if (mysql_num_rows($sqls)!=0) {
while ($row = mysql_fetch_assoc($sqls)){
$weight = $row["weight"];
echo "C" . $weight;
}}
else {
echo "No stats found";
}
?>