Php 从命令行传递给函数时不处理值 当我将变量传递给我从命令行获得的函数时,它返回无效
在validate_vendor函数中,我嵌入了给定bwlow的逻辑,即如果vendor name不在这些vendor类型中,则返回0,否则返回1。即使我按下有效的供应商名称,如“vendor1”,它也会给出无效的供应商名称并退出Php 从命令行传递给函数时不处理值 当我将变量传递给我从命令行获得的函数时,它返回无效,php,command-line,Php,Command Line,在validate_vendor函数中,我嵌入了给定bwlow的逻辑,即如果vendor name不在这些vendor类型中,则返回0,否则返回1。即使我按下有效的供应商名称,如“vendor1”,它也会给出无效的供应商名称并退出 function validate_vendor($prev_vendor, &$tables) { if($prev_vendor == "vendor1") { ---
function validate_vendor($prev_vendor, &$tables) {
if($prev_vendor == "vendor1")
{
---
---
return 1;
}
if($prev_vendor == "vendor2")
{
---
---
return 1;
}
if($prev_vendor == "vendor3")
{
---
---
return 1;
}
if($prev_vendor == "vendor4")
{
---
---
return 1;
}
if($prev_vendor == "vendor5")
{
---
---
return 1;
}
else
{
return 0;
}
}
你做过调试吗?第一步肯定是检查从
stdin
获得的$vendor
值,对吗?它包括你点击return时的新行吗?来吧,努力,调试一下!:-)好的,thanx,经过研究我找到了解决方案
function validate_vendor($prev_vendor, &$tables) {
if($prev_vendor == "vendor1")
{
---
---
return 1;
}
if($prev_vendor == "vendor2")
{
---
---
return 1;
}
if($prev_vendor == "vendor3")
{
---
---
return 1;
}
if($prev_vendor == "vendor4")
{
---
---
return 1;
}
if($prev_vendor == "vendor5")
{
---
---
return 1;
}
else
{
return 0;
}
}