问题
Why does the following code output 0?
It works with numbers instead of strings just fine. I have similar code in JavaScript that also works. Does PHP not like += with strings?
<?php
$selectBox = '<select name="number">';
for ($i=1; $i<=100; $i++)
{
$selectBox += '<option value="' . $i . '">' . $i . '</option>';
}
$selectBox += '</select>';
echo $selectBox;
?>
回答1:
This is because PHP uses the period character .
for string concatenation, not the plus character +
. Therefore to append to a string you want to use the .=
operator:
for ($i=1;$i<=100;$i++)
{
$selectBox .= '<option value="' . $i . '">' . $i . '</option>';
}
$selectBox .= '</select>';
回答2:
In PHP use .=
to append strings, and not +=
.
Why does this output 0? [...] Does PHP not like += with strings?
+=
is an arithmetic operator to add a number to another number. Using that operator with strings leads to an automatic type conversion. In the OP's case the strings have been converted to integers of the value 0
.
More about operators in PHP:
- Reference - What does this symbol mean in PHP?
- PHP Manual – Operators
回答3:
PHP syntax is little different in case of concatenation from JavaScript.
Instead of (+) plus
a (.) period
is used for string concatenation.
<?php
$selectBox = '<select name="number">';
for ($i=1;$i<=100;$i++)
{
$selectBox += '<option value="' . $i . '">' . $i . '</option>'; // <-- (Wrong) Replace + with .
$selectBox .= '<option value="' . $i . '">' . $i . '</option>'; // <-- (Correct) Here + is replaced .
}
$selectBox += '</select>'; // <-- (Wrong) Replace + with .
$selectBox .= '</select>'; // <-- (Correct) Here + is replaced .
echo $selectBox;
?>
来源:https://stackoverflow.com/questions/64740320/convert-sql-query-with-coalesce-and-left-join-to-php