It is estimated that it is a punctuation problem. I can't find it. I want to help you.

It is estimated that it is a punctuation problem. I can't find it. I want to help you.

for($i=0;$i<$num;++$i){    $ming=$a1[$i];    $zjming=$a2[$i];    $zjtime=$a3[$i];    $mysqli->query('update xs set xinzhangjie='$zjming',zhangjieshijian='$zjtime' where xsMing='$ming'');     printf("Affected rows (UPDATE): %d ", $mysqli->affected_rows);    echo $ming.'---';

MYSQLI returns-1, indicating that the execution department is successful.
You can update a single-pick statement.
All variables can be recycled out of values. It is loop update, and it won't work. it is estimated that it is the punctuation of the update statement. I have tested many types of punctuation.
For example.

Xinzhangjie = '". $ zjming."' No
Xinzhangjie = '{$ zjming}' no
Xinzhangjie = "$ zjming" no
Xinzhangjie = ". $ zjming." No

..... My head is big.

Reply to discussion (solution)

$ Mysqli-> query ("update xs set xinzhangjie = '$ zjming', zhangjieshijian = '$ zjtime' where xsMing =' $ Ming '");

Double quotation marks

$ Mysqli-> query ("update xs set xinzhangjie = '$ zjming', zhangjieshijian = '$ zjtime' where xsMing =' $ Ming '");


Thank you so much, bro.

Double quotation marks

OK. Thank you for your help.

Use a hyphen to link strings and variables ..
The connection character is "."

Single quotes do not parse PHP variables...

So...

Single quotes do not parse PHP variables...

So...


Use a hyphen to link strings and variables ..
The connection character is "."

Therefore, double quotation marks are used. Now I understand. Thank you !~!~!~