我想实现此查询:
if(x=1){
$update = "close = '$date'";
}
else {
$update = "open = '$date'";
}
$query = "Update table1 set $update where id=100";
mysql_query($query);
但我收到错误,Mysql无法执行查询?
答案 0 :(得分:1)
<?php
if($x==1){
$update = "close = '".$date."'";
}
else {
$update = "open = '".$date."'";
}
$query = "update table1 set $update where id=100";
mysql_query($query);
?>
答案 1 :(得分:0)
将您的查询放在引号中。请尝试以下方法:
if(x==1){
$update = "close='".$date."'";
}
else {
$update = "open = '".$date."'";
}
$query = "Update table1 set ".$update." where id=100";
mysql_query($query) or die(mysql_error());
使用mysql_error()来检查你从mysql获得的错误
答案 2 :(得分:0)
使用此
<?php
if(x==1){
$update = "close = '$date'";
}
else {
$update = "open = '$date'";
}
$query = "Update table1 set '".$update."' where id=100";
mysql_query($query);
?>
答案 3 :(得分:0)
尝试使用此代码........
<?php
if($x==1){ // use $ for variables
$update = "close = '".$date."' "; //always concatenate variables
}
else {
$update = "open = '".$date."' ";
}
$query = "Update table1 set '".$update."' where id=100";
mysql_query($query, $connection); // don`t forget to add mysql connection
?>
答案 4 :(得分:-1)
Replace
$query = Update table1 set $update where id=100;
to
$query = "Update table1 set ".$update." where id=100";