本文介绍了未定义的偏移量:的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
$result = db2_getsub( array('campaigns_id' => $this_id) );
if (!is_null($result))
{
$numsub = 0;
while ( $row = $result->fetch_object() )
{
$numsub = $numsub +1;
array_push($arrFornavn, $row->sub_name);
array_push($arrEtternavn, $row->sub_code);
}
$formcountfld=$numsub;
}
else
{
$numsub=1;
$formcountfld=1;
}
$i = 1;
if (1==1) {
for ($i = 1; $i <= $numsub; $i++)
{
?>
<div class="fieldrow_horz">
<div class="fieldgroup">
<input type="text" id="fornavn_<?= $i ?>" name="fornavn_<?= $i ?>" value="<?= $arrFornavn[$i-1] ?>" />
</div>
<div class="fieldgroup">
<input type="text" id="etternavn_<?= $i ?>" name="etternavn_<?= $i ?>" value="<?= $arrEtternavn[$i-1] ?>" />
</div>
</div>
<?php
}
} else {
?>
我收到的错误提示:未定义偏移量:0
任何人都可以帮助。在此先感谢您的期待。
I am getting an error as Notice: Undefined offset: 0
can anybody help .thanks in advance looking forward for an answer.
推荐答案
这意味着您正在尝试访问一个不存在的值,在您的情况下为$ arrEtternavn [0]。
It means you are trying to access a value that does not exist, which in your case is $arrEtternavn[0].
小例子:
$array = array();
$array[1] = 'one';
$array[2] = 'two';
$array[4] = 'four';
echo $array[0]; // This will give a notice, $array[0] does not exist.
echo $array[1]; // 'one'
echo $array[2]; // 'two'
echo $array[3]; // This will give a notice, $array[3] does not exist.
echo $array[4]; // 'four'
一个非常简单的解决方法就是:
A very easy fix would me something like this:
value="<?= isset($arrEtternavn[$i-1]) ? $arrEtternavn[$i-1] : '' ?>"
这是一个简短的elseif->(条件)? if_True:if_False
This is a short elseif-> (condition) ? if_True : if_False
编辑:我想补充一点,通知并不是一件坏事。没有通知会是最好的事情,但它不应该让您在晚上醒着。
I'd like to add that a notice isnt a very bad thing. No notices would be the best thing, but it should not keep you awake at night.
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